Chapter 1: Real Numbers

Solution

By definition, the remainder r satisfies 0 ≤ r < b. It can be 0 (exact division) but must be strictly less than the divisor b.

Solution

3825 ÷ 3 = 1275; 1275 ÷ 3 = 425; 425 ÷ 5 = 85; 85 ÷ 5 = 17 (prime).

3825 = 3² × 5² × 17

Solution

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

HCF = 51

Solution

12576 = 4052 × 3 + 420

4052 = 420 × 9 + 272

420 = 272 × 1 + 148

272 = 148 × 1 + 124

148 = 124 × 1 + 24

124 = 24 × 5 + 4

24 = 4 × 6 + 0

HCF = 4

Solution

306 = 2 × 3² × 17; 657 = 3² × 73

HCF = 3² = 9; LCM = 2 × 3² × 17 × 73 = 22338

9 × 22338 = 201042 = 306 × 657 ✓

Solution

3125 = 5⁵ (form 2ⁿ5ᵐ) → terminating. Others have prime factors besides 2 and 5.

Solution

8 = 2³. Multiply: 17×5³/(2³×5³) = 17×125/1000 = 2125/1000 = 2.125

Solution

HCF × LCM = a × b → 12 × LCM = 1800 → LCM = 150

Solution

√4=2, √18/√2=3, √12×√3=6 — all rational. √5−3 is irrational (irrational minus rational).

Solution

33/20 = 165/100 = 1.65 → 2 decimal places. Rule: max(n,m) where denom = 2ⁿ×5ᵐ → max(2,1)=2.

Solution

Using Euclid's division algorithm to find HCF of 119 and 178:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(119, 178) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 249 and 120:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(249, 120) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 54 and 258:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(54, 258) = 6\)

Solution

Using Euclid's division algorithm to find HCF of 257 and 285:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(257, 285) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 93 and 187:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(93, 187) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 233 and 272:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(233, 272) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 221 and 257:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(221, 257) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 66 and 69:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(66, 69) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 24 and 257:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(24, 257) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 70 and 69:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(70, 69) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 36 and 83:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(36, 83) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 210 and 119:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(210, 119) = 7\)

Solution

Using Euclid's division algorithm to find HCF of 215 and 52:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(215, 52) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 45 and 94:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(45, 94) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 234 and 136:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(234, 136) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 96 and 220:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(96, 220) = 4\)

Solution

Using Euclid's division algorithm to find HCF of 37 and 131:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(37, 131) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 101 and 35:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(101, 35) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 283 and 226:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(283, 226) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 241 and 58:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(241, 58) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 114 and 80:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(114, 80) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 65 and 150:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(65, 150) = 5\)

Solution

Using Euclid's division algorithm to find HCF of 167 and 261:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(167, 261) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 187 and 54:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(187, 54) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 107 and 78:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(107, 78) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 181 and 54:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(181, 54) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 96 and 248:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(96, 248) = 8\)

Solution

Using Euclid's division algorithm to find HCF of 146 and 254:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(146, 254) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 87 and 282:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(87, 282) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 48 and 164:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(48, 164) = 4\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(52,7) = 364\) ... wait, let's compute:

\(\text{LCM}(52,7) = \frac{52 \times 7}{\text{HCF}(52,7)} = 364\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(10,35) = 70\) ... wait, let's compute:

\(\text{LCM}(10,35) = \frac{10 \times 35}{\text{HCF}(10,35)} = 70\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(42,33) = 462\) ... wait, let's compute:

\(\text{LCM}(42,33) = \frac{42 \times 33}{\text{HCF}(42,33)} = 462\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(37,56) = 2072\) ... wait, let's compute:

\(\text{LCM}(37,56) = \frac{37 \times 56}{\text{HCF}(37,56)} = 2072\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(12,17) = 204\) ... wait, let's compute:

\(\text{LCM}(12,17) = \frac{12 \times 17}{\text{HCF}(12,17)} = 204\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(49,59) = 2891\) ... wait, let's compute:

\(\text{LCM}(49,59) = \frac{49 \times 59}{\text{HCF}(49,59)} = 2891\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(31,45) = 1395\) ... wait, let's compute:

\(\text{LCM}(31,45) = \frac{31 \times 45}{\text{HCF}(31,45)} = 1395\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(53,6) = 318\) ... wait, let's compute:

\(\text{LCM}(53,6) = \frac{53 \times 6}{\text{HCF}(53,6)} = 318\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(10,36) = 180\) ... wait, let's compute:

\(\text{LCM}(10,36) = \frac{10 \times 36}{\text{HCF}(10,36)} = 180\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(26,26) = 26\) ... wait, let's compute:

\(\text{LCM}(26,26) = \frac{26 \times 26}{\text{HCF}(26,26)} = 26\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(32,38) = 608\) ... wait, let's compute:

\(\text{LCM}(32,38) = \frac{32 \times 38}{\text{HCF}(32,38)} = 608\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(55,49) = 2695\) ... wait, let's compute:

\(\text{LCM}(55,49) = \frac{55 \times 49}{\text{HCF}(55,49)} = 2695\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(6,12) = 12\) ... wait, let's compute:

\(\text{LCM}(6,12) = \frac{6 \times 12}{\text{HCF}(6,12)} = 12\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(18,12) = 36\) ... wait, let's compute:

\(\text{LCM}(18,12) = \frac{18 \times 12}{\text{HCF}(18,12)} = 36\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(19,15) = 285\) ... wait, let's compute:

\(\text{LCM}(19,15) = \frac{19 \times 15}{\text{HCF}(19,15)} = 285\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(31,22) = 682\) ... wait, let's compute:

\(\text{LCM}(31,22) = \frac{31 \times 22}{\text{HCF}(31,22)} = 682\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(28,39) = 1092\) ... wait, let's compute:

\(\text{LCM}(28,39) = \frac{28 \times 39}{\text{HCF}(28,39)} = 1092\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(40,38) = 760\) ... wait, let's compute:

\(\text{LCM}(40,38) = \frac{40 \times 38}{\text{HCF}(40,38)} = 760\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(47,17) = 799\) ... wait, let's compute:

\(\text{LCM}(47,17) = \frac{47 \times 17}{\text{HCF}(47,17)} = 799\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(45,10) = 90\) ... wait, let's compute:

\(\text{LCM}(45,10) = \frac{45 \times 10}{\text{HCF}(45,10)} = 90\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(42,26) = 546\) ... wait, let's compute:

\(\text{LCM}(42,26) = \frac{42 \times 26}{\text{HCF}(42,26)} = 546\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(20,55) = 220\) ... wait, let's compute:

\(\text{LCM}(20,55) = \frac{20 \times 55}{\text{HCF}(20,55)} = 220\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(27,44) = 1188\) ... wait, let's compute:

\(\text{LCM}(27,44) = \frac{27 \times 44}{\text{HCF}(27,44)} = 1188\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(11,28) = 308\) ... wait, let's compute:

\(\text{LCM}(11,28) = \frac{11 \times 28}{\text{HCF}(11,28)} = 308\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(24,38) = 456\) ... wait, let's compute:

\(\text{LCM}(24,38) = \frac{24 \times 38}{\text{HCF}(24,38)} = 456\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(6,59) = 354\) ... wait, let's compute:

\(\text{LCM}(6,59) = \frac{6 \times 59}{\text{HCF}(6,59)} = 354\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(57,23) = 1311\) ... wait, let's compute:

\(\text{LCM}(57,23) = \frac{57 \times 23}{\text{HCF}(57,23)} = 1311\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(46,18) = 414\) ... wait, let's compute:

\(\text{LCM}(46,18) = \frac{46 \times 18}{\text{HCF}(46,18)} = 414\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(39,54) = 702\) ... wait, let's compute:

\(\text{LCM}(39,54) = \frac{39 \times 54}{\text{HCF}(39,54)} = 702\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(10,26) = 130\) ... wait, let's compute:

\(\text{LCM}(10,26) = \frac{10 \times 26}{\text{HCF}(10,26)} = 130\)

Solution

Using Euclid's division algorithm to find HCF of 80 and 30:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(80, 30) = 10\)

Solution

Using Euclid's division algorithm to find HCF of 148 and 216:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(148, 216) = 4\)

Solution

Using Euclid's division algorithm to find HCF of 49 and 58:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(49, 58) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 169 and 65:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(169, 65) = 13\)

Solution

Using Euclid's division algorithm to find HCF of 206 and 91:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(206, 91) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 160 and 287:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(160, 287) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 198 and 177:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(198, 177) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 113 and 238:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(113, 238) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 130 and 56:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(130, 56) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 39 and 59:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(39, 59) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 213 and 198:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(213, 198) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 112 and 115:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(112, 115) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 222 and 217:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(222, 217) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 24 and 228:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(24, 228) = 12\)

Solution

Using Euclid's division algorithm to find HCF of 274 and 39:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(274, 39) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 196 and 220:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(196, 220) = 4\)

Solution

Using Euclid's division algorithm to find HCF of 55 and 61:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(55, 61) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 224 and 126:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(224, 126) = 14\)

Solution

Using Euclid's division algorithm to find HCF of 29 and 90:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(29, 90) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 132 and 111:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(132, 111) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 182 and 255:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(182, 255) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 294 and 79:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(294, 79) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 46 and 178:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(46, 178) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 238 and 155:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(238, 155) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 112 and 74:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(112, 74) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 212 and 173:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(212, 173) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 249 and 56:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(249, 56) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 249 and 129:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(249, 129) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 186 and 297:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(186, 297) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 113 and 283:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(113, 283) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 296 and 192:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(296, 192) = 8\)

Solution

Using Euclid's division algorithm to find HCF of 78 and 80:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(78, 80) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 39 and 233:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(39, 233) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 281 and 256:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(281, 256) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 249 and 27:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(249, 27) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 28 and 43:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(28, 43) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 149 and 197:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(149, 197) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 58 and 198:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(58, 198) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 54 and 276:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(54, 276) = 6\)

Solution

Using Euclid's division algorithm to find HCF of 91 and 299:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(91, 299) = 13\)

Solution

Using Euclid's division algorithm to find HCF of 156 and 123:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(156, 123) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 57 and 156:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(57, 156) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 48 and 179:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(48, 179) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 83 and 216:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(83, 216) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 74 and 150:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(74, 150) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 198 and 244:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(198, 244) = 2\)

Solution

Using Euclid's division algorithm to find HCF of 274 and 281:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(274, 281) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 40 and 49:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(40, 49) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 270 and 267:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(270, 267) = 3\)

Solution

Using Euclid's division algorithm to find HCF of 166 and 49:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(166, 49) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 249 and 80:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(249, 80) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 255 and 271:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(255, 271) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 51 and 253:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(51, 253) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 84 and 98:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(84, 98) = 14\)

Solution

Using Euclid's division algorithm to find HCF of 159 and 74:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(159, 74) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 162 and 288:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(162, 288) = 18\)

Solution

Using Euclid's division algorithm to find HCF of 25 and 251:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(25, 251) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 95 and 81:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(95, 81) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 239 and 196:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(239, 196) = 1\)

Solution

Using Euclid's division algorithm to find HCF of 241 and 104:

We repeatedly divide the larger by the smaller and take remainders.

\(\text{HCF}(241, 104) = 1\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(46,11) = 506\) ... wait, let's compute:

\(\text{LCM}(46,11) = \frac{46 \times 11}{\text{HCF}(46,11)} = 506\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(12,58) = 348\) ... wait, let's compute:

\(\text{LCM}(12,58) = \frac{12 \times 58}{\text{HCF}(12,58)} = 348\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(55,29) = 1595\) ... wait, let's compute:

\(\text{LCM}(55,29) = \frac{55 \times 29}{\text{HCF}(55,29)} = 1595\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(46,17) = 782\) ... wait, let's compute:

\(\text{LCM}(46,17) = \frac{46 \times 17}{\text{HCF}(46,17)} = 782\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(14,40) = 280\) ... wait, let's compute:

\(\text{LCM}(14,40) = \frac{14 \times 40}{\text{HCF}(14,40)} = 280\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(12,48) = 48\) ... wait, let's compute:

\(\text{LCM}(12,48) = \frac{12 \times 48}{\text{HCF}(12,48)} = 48\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(49,23) = 1127\) ... wait, let's compute:

\(\text{LCM}(49,23) = \frac{49 \times 23}{\text{HCF}(49,23)} = 1127\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(38,34) = 646\) ... wait, let's compute:

\(\text{LCM}(38,34) = \frac{38 \times 34}{\text{HCF}(38,34)} = 646\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(20,8) = 40\) ... wait, let's compute:

\(\text{LCM}(20,8) = \frac{20 \times 8}{\text{HCF}(20,8)} = 40\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(42,53) = 2226\) ... wait, let's compute:

\(\text{LCM}(42,53) = \frac{42 \times 53}{\text{HCF}(42,53)} = 2226\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(26,46) = 598\) ... wait, let's compute:

\(\text{LCM}(26,46) = \frac{26 \times 46}{\text{HCF}(26,46)} = 598\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(58,49) = 2842\) ... wait, let's compute:

\(\text{LCM}(58,49) = \frac{58 \times 49}{\text{HCF}(58,49)} = 2842\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(6,53) = 318\) ... wait, let's compute:

\(\text{LCM}(6,53) = \frac{6 \times 53}{\text{HCF}(6,53)} = 318\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(37,12) = 444\) ... wait, let's compute:

\(\text{LCM}(37,12) = \frac{37 \times 12}{\text{HCF}(37,12)} = 444\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(29,6) = 174\) ... wait, let's compute:

\(\text{LCM}(29,6) = \frac{29 \times 6}{\text{HCF}(29,6)} = 174\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(60,48) = 240\) ... wait, let's compute:

\(\text{LCM}(60,48) = \frac{60 \times 48}{\text{HCF}(60,48)} = 240\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(29,33) = 957\) ... wait, let's compute:

\(\text{LCM}(29,33) = \frac{29 \times 33}{\text{HCF}(29,33)} = 957\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(19,33) = 627\) ... wait, let's compute:

\(\text{LCM}(19,33) = \frac{19 \times 33}{\text{HCF}(19,33)} = 627\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(18,47) = 846\) ... wait, let's compute:

\(\text{LCM}(18,47) = \frac{18 \times 47}{\text{HCF}(18,47)} = 846\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(52,33) = 1716\) ... wait, let's compute:

\(\text{LCM}(52,33) = \frac{52 \times 33}{\text{HCF}(52,33)} = 1716\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(9,58) = 522\) ... wait, let's compute:

\(\text{LCM}(9,58) = \frac{9 \times 58}{\text{HCF}(9,58)} = 522\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(22,33) = 66\) ... wait, let's compute:

\(\text{LCM}(22,33) = \frac{22 \times 33}{\text{HCF}(22,33)} = 66\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(35,29) = 1015\) ... wait, let's compute:

\(\text{LCM}(35,29) = \frac{35 \times 29}{\text{HCF}(35,29)} = 1015\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(34,42) = 714\) ... wait, let's compute:

\(\text{LCM}(34,42) = \frac{34 \times 42}{\text{HCF}(34,42)} = 714\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(10,24) = 120\) ... wait, let's compute:

\(\text{LCM}(10,24) = \frac{10 \times 24}{\text{HCF}(10,24)} = 120\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(57,25) = 1425\) ... wait, let's compute:

\(\text{LCM}(57,25) = \frac{57 \times 25}{\text{HCF}(57,25)} = 1425\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(18,17) = 306\) ... wait, let's compute:

\(\text{LCM}(18,17) = \frac{18 \times 17}{\text{HCF}(18,17)} = 306\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(7,32) = 224\) ... wait, let's compute:

\(\text{LCM}(7,32) = \frac{7 \times 32}{\text{HCF}(7,32)} = 224\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(37,27) = 999\) ... wait, let's compute:

\(\text{LCM}(37,27) = \frac{37 \times 27}{\text{HCF}(37,27)} = 999\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(29,25) = 725\) ... wait, let's compute:

\(\text{LCM}(29,25) = \frac{29 \times 25}{\text{HCF}(29,25)} = 725\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(29,23) = 667\) ... wait, let's compute:

\(\text{LCM}(29,23) = \frac{29 \times 23}{\text{HCF}(29,23)} = 667\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(46,54) = 1242\) ... wait, let's compute:

\(\text{LCM}(46,54) = \frac{46 \times 54}{\text{HCF}(46,54)} = 1242\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(26,7) = 182\) ... wait, let's compute:

\(\text{LCM}(26,7) = \frac{26 \times 7}{\text{HCF}(26,7)} = 182\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(32,32) = 32\) ... wait, let's compute:

\(\text{LCM}(32,32) = \frac{32 \times 32}{\text{HCF}(32,32)} = 32\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(50,7) = 350\) ... wait, let's compute:

\(\text{LCM}(50,7) = \frac{50 \times 7}{\text{HCF}(50,7)} = 350\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(23,44) = 1012\) ... wait, let's compute:

\(\text{LCM}(23,44) = \frac{23 \times 44}{\text{HCF}(23,44)} = 1012\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(20,31) = 620\) ... wait, let's compute:

\(\text{LCM}(20,31) = \frac{20 \times 31}{\text{HCF}(20,31)} = 620\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(48,56) = 336\) ... wait, let's compute:

\(\text{LCM}(48,56) = \frac{48 \times 56}{\text{HCF}(48,56)} = 336\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(9,19) = 171\) ... wait, let's compute:

\(\text{LCM}(9,19) = \frac{9 \times 19}{\text{HCF}(9,19)} = 171\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(45,8) = 360\) ... wait, let's compute:

\(\text{LCM}(45,8) = \frac{45 \times 8}{\text{HCF}(45,8)} = 360\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(31,57) = 1767\) ... wait, let's compute:

\(\text{LCM}(31,57) = \frac{31 \times 57}{\text{HCF}(31,57)} = 1767\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(29,19) = 551\) ... wait, let's compute:

\(\text{LCM}(29,19) = \frac{29 \times 19}{\text{HCF}(29,19)} = 551\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(6,7) = 42\) ... wait, let's compute:

\(\text{LCM}(6,7) = \frac{6 \times 7}{\text{HCF}(6,7)} = 42\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(24,30) = 120\) ... wait, let's compute:

\(\text{LCM}(24,30) = \frac{24 \times 30}{\text{HCF}(24,30)} = 120\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(45,36) = 180\) ... wait, let's compute:

\(\text{LCM}(45,36) = \frac{45 \times 36}{\text{HCF}(45,36)} = 180\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(14,27) = 378\) ... wait, let's compute:

\(\text{LCM}(14,27) = \frac{14 \times 27}{\text{HCF}(14,27)} = 378\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(20,41) = 820\) ... wait, let's compute:

\(\text{LCM}(20,41) = \frac{20 \times 41}{\text{HCF}(20,41)} = 820\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(20,48) = 240\) ... wait, let's compute:

\(\text{LCM}(20,48) = \frac{20 \times 48}{\text{HCF}(20,48)} = 240\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(19,36) = 684\) ... wait, let's compute:

\(\text{LCM}(19,36) = \frac{19 \times 36}{\text{HCF}(19,36)} = 684\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(9,57) = 171\) ... wait, let's compute:

\(\text{LCM}(9,57) = \frac{9 \times 57}{\text{HCF}(9,57)} = 171\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(52,17) = 884\) ... wait, let's compute:

\(\text{LCM}(52,17) = \frac{52 \times 17}{\text{HCF}(52,17)} = 884\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(60,21) = 420\) ... wait, let's compute:

\(\text{LCM}(60,21) = \frac{60 \times 21}{\text{HCF}(60,21)} = 420\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(31,13) = 403\) ... wait, let's compute:

\(\text{LCM}(31,13) = \frac{31 \times 13}{\text{HCF}(31,13)} = 403\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(28,38) = 532\) ... wait, let's compute:

\(\text{LCM}(28,38) = \frac{28 \times 38}{\text{HCF}(28,38)} = 532\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(53,44) = 2332\) ... wait, let's compute:

\(\text{LCM}(53,44) = \frac{53 \times 44}{\text{HCF}(53,44)} = 2332\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(39,43) = 1677\) ... wait, let's compute:

\(\text{LCM}(39,43) = \frac{39 \times 43}{\text{HCF}(39,43)} = 1677\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(55,55) = 55\) ... wait, let's compute:

\(\text{LCM}(55,55) = \frac{55 \times 55}{\text{HCF}(55,55)} = 55\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(55,45) = 495\) ... wait, let's compute:

\(\text{LCM}(55,45) = \frac{55 \times 45}{\text{HCF}(55,45)} = 495\)

Solution

We know \(\text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\)

\(\text{HCF}(14,38) = 266\) ... wait, let's compute:

\(\text{LCM}(14,38) = \frac{14 \times 38}{\text{HCF}(14,38)} = 266\)

Solution

Required number divides (245−5)=240 and (1029−5)=1024.

240 = 2⁴×3×5; 1024 = 2¹⁰

HCF = 2⁴ = 16

Solution

Assume √5 = a/b (co-prime). Then a² = 5b², so 5|a, let a = 5k. Then 25k² = 5b² → b² = 5k², so 5|b. Both divisible by 5 contradicts co-primality. Hence √5 is irrational.

Solution

Assume 3 + 2√5 = a/b (rational). Then √5 = (a−3b)/(2b), which is rational — contradicts the irrationality of √5. Hence 3 + 2√5 is irrational.

Solution

9 × 360 = 45 × x → x = 3240/45 = 72

Solution

LCM(15,20,36) = 2²×3²×5 = 180. Required = 180 + 5 = 185

Solution

a = 6q+r, r ∈ {0,1,2,3,4,5}. Even values: 6q, 6q+2, 6q+4. So odd integers: 6q+1, 6q+3, 6q+5.

Solution

6ⁿ = 2ⁿ×3ⁿ has no factor 5, so never divisible by 10. Cannot end in 0. (In fact 6ⁿ always ends in 6.)

Solution

LCM(9,12,15) = 2²×3²×5 = 180 min = 3 hours. Next: 8 AM + 3 h = 11:00 AM.

Solution

Assume √2+√3 = r (rational). Then √3 = r−√2. Square: 3 = r²−2r√2+2 → √2 = (r²−1)/(2r), rational — contradiction. Hence irrational.

Solution

a = 3q → a² = 9q² = 3(3q²) = 3m.

a = 3q+1 → a² = 9q²+6q+1 = 3(3q²+2q)+1 = 3m+1.

a = 3q+2 → a² = 9q²+12q+4 = 3(3q²+4q+1)+1 = 3m+1.