Find the smallest number which when increased by 17

\[520 = 2^3 \times 5 \times 13\] `468 = 2^2xx3^2xx13` \[\text = 2^3 \times 3^2 \times 5 \times 13\] \[ = 4680\] Hence 4680 is the least number which exactly divides 520 and 468 i.e. we will get a remainder of 0 in this case. But we need the Smallest number which when increased by 17 is exactly divided by 520 and 468. Therefore = 4680 -17 = 4663 Hence 4663 is Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Dear student, The given numbers are 520 and 468. The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468. Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13 Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13 LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680.(this is obtained by first taking terms common in both factorisation of 520 and 468(here 2×2×13 is common, hence we write, 2×2×13.) Now write the remaining numbers i.e., 2x5 in 520 and 3x3 in 468). Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680-17 = 4663. Thank you.

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