Find the smallest number which when increased by 17
Question:

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Solution:

TO FIND: Smallest number which when increased by 17 is exactly divisible by both 520 and 468.

L.C.M OF 520 and 468

$520=2^{3} \times 5 \times 13$

$468=2^{2} \times 3^{2} \times 13$

LCM of 520 and $468=2^{3} \times 3^{2} \times 5 \times 13$

= 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 .

Administrator

Leave a comment

Please enter comment.
Please enter your name.