Question

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

Answer 1

Small number divisible by 520 and 468 when it is increased by 17 is 4663

Solution:

To find smallest number dividing both 520 and 468 when increased by 17, can be found out by LCM method

$\begin{array}{l}{520=2 \times 2 \times 2 \times 5 \times 13} \\ {468=2 \times 2 \times 3 \times 3 \times 13}\end{array}$

LCM of 520 and 468 = $2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 13=4680$

Now the total value that is to be divided or the smallest number after increase by 17 becomes 4680, therefore the smallest number will be,

4680 – 17 = 4663

Therefore the number after subtraction of 17 becomes 4663  

Answer 2

Small number divisible by both 520 and 468 when it is increased by 17 is 4663.

Solution:

Given: The given numbers are 520 and 468.

In order to find out the smallest number which, we will have to find out the LCM of 520 and 468 and subtract 17 from that L.C.M.

Factors of $520=2 \times 2 \times 2 \times 5 \times 13$

Factors of $468=2 \times 2 \times 3 \times 3 \times 13$

$\mathrm{LCM}=2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 13=4680$

Therefore, the required smallest number  

= 4680 – 17  

= 4663

Hence, when 4663 is increased by 17, we get 4680 which is the L.C.M. of 520 and 468 i.e. is divisible by both.