State true or false: The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.

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True

False

answer is A.

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Detailed Solution

Here, we have to decide if the given statement i.e., The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663 is true or not.
As per the statement, we have to find the smallest number which is divisible by 468 and 520, when increased by 17, which will be nothing but the LCM of 468 and 520.
So, we proceed with the prime factorization of 468 and 520 and then find the LCM as follows,

520 = 2 × 2 × 2 × 5 × 13 468 = 2 × 2 × 3 × 3 × 13

LCM (520,468)

= 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680

Now, as mentioned in the statement, the LCM is increased by 17, so the actual number which is divisible by 468 and 520 must be reduced by 17.
Therefore, the required number will be

4680 − 17 = 4663

.
Hence, the statement is true.
So, the correct option is 1.

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