Find the smallest number when divided by 28, 40 and 44 leave a remainder 8 in each case.

To find: Smallest number when divided by 28,40 and 44 leave a remainder 8 in each case.

Solution:

Taking prime factorization of 28, 40 and 44:

$28 = 2^2 \times 7$

$40 = 2^3 \times 5$

To find LCM, we have multiply the prime factors along with maximum powers

$LCM = 2^3 \times 5 \times 7 \times 11$

LCM = 3080

Since 3080 is LCM of 28, 40 and 44. It is the smallest number that divides all the three without leaving any remainder. 

$3080+8 = 3088$

3088 is the smallest number when divided by 28,40 and 44 leaves a remainder 8

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Updated on: 10-Oct-2022

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