Question

Find the smallest number which leaves remainders 2, 6 and 12 when divided by 22, 26 and 32 respectively please send me important​

Answer 1

Here (22-2)=20, (26-6)=20, (32-12) = 20.

.°. required no. = (LCM 22,26,32)-20

22 = 2×11

26 = 2×13

32 = 2×2×2×2×2 = 2^5

LCM = highest power of all factors

= 2^5 ×11×13 = 4576

required no. = 4576-20 = 4556

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