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Question
find the smallest number which when increased by 17 is exactly divisible by 420 and 468
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Solution
The given numbers are 420 and 468.
The smallest number which when increased by 17 is exactly divisible by both 420 and 468 is obtained by subtracting 17 from the LCM of 420 and 468.
Prime factorisation of 420 = 2 × 2 × 3 × 5 ×7
Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13
LCM of 520 and 468 = 2 × 2 × 3 × 3 × 5 ×7 ×13 = 16380.
Smallest number which when increased by 17 is exactly divisible by both 420 and 468 =16380. – 17 = 16363.
The given numbers are 420 and 468.
The smallest number which when increased by 17 is exactly divisible by both 420 and 468 is obtained by subtracting 17 from the LCM of 420 and 468.
Prime factorisation of 420 = 2 × 2 × 3 × 5 ×7
Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13
LCM of 520 and 468 = 2 × 2 × 3 × 3 × 5 ×7 ×13 = 16380.
Smallest number which when increased by 17 is exactly divisible by both 420 and 468 =16380. – 17 = 16363.
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