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Quantitative Aptitude

HCF and LCM

Find the leas...

Question

Find the least number which when divided by 2, 3, 4, 5 and 6 leaves a reminder of 1 in each case but when divided by 7 leaves no remainder.

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Solution

Therefore the required number is of the form 60x + 1 (since it leaves a reminder of 1)
The number is a multiple of 7.
60x + 1 is a multiple of 7
56x + 4x + 1 is a multiple of 7
The least value of x for which this holds true is 12.
60x+ 1 = 60(12) + 1 = 721

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