find the smallest number which when incresed by 17is exactly divisible both 520 and .468
Answers
Answered by
4
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. Smallest number which when increased by 17 is exactly divisible by both 520 and 468 =4680 – 17 = 4663.
aayat45:
thanks
plz mark as brain list
but I think to find smallest we should find HCF and subtract 17 from it
Similar questions