Aptitude - Problems on H.C.F and L.C.M - Discussion

Where we meet the condition to be a multiple of 7?

5*1 = 5.
2*3 = 6.
2*2 = 4.
3*1 = 3.

5*1*2*3*2*2*3*1.
5*2*3*2 = 60.

2497/60 = Remainder 37.
60-37 = 23.

Nos is 2497 opt 3, 13, 23, 33 which divide 3, 4, 5, 6.

Add each one:

2497+3 = 2500.
2497+13 = 2510.
2497+23 = 2520.
2497+33 = 2530.

3 divides only when addition of those no is divisible by 3.

So 2500 = 7 not divide.

2510 = 8
2520 = 9 divide.

Can anyone explain how the sum is exactly divisible by 5, 6, 4 and 3?

Sum of 23 is divisible only by 5 right?

Please explain by equation 60-37?

Guys please see option and each option on number and using divisibility rule. Check if it is perfectly divisible or not.

Sol: LCM of 5,6,4,3 is 60.

On dividing 2497 by 60 we get 37 as remainder.

Therefore number to added is 60 - 37 =23.

Now accordingly we get lcm of 5,4,6 and 3 as 20. Now as it says that when wats the least number to be added with 2497 so that it get divided by 60 completely.

Now when we divide 2497 by 60 we get remainder as 37(left over). Now 37 would not be the remainder or left over and be divided by 60 completely if 33 more was there 23 more was there with 37(23+37=60).

So we got 23 by equation calculating 60-37 as that is the only amount needed to add with 37 so that its divisible by 60.

So in other words 23 is the only amount to be added to 2497 to get it fully divided by 60(2497+23).

What would have been the case if we are asked to find the highest no. Which should be added to 2497?

Then are we supposed to add 37 to 2497 ?

Why 60-37 please explain?