Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Q.
Therefore, 4663 is the required answer.
The least number divisible by both 520 and 468 is the
Least Common Multiple of 520 and 468(L.C.M)
Sol.
520
5
=
2
×
2
×
2
×
13
×
468
3
=
2
×
2
×
3
×
13
×
Clearly,
=
L.C.M.
13
23
×
32
×
5
×
= 4680
So, least number divisible by both 520 and 468 is 4680.
Let the number which when increased by 17 gives the least number divisible by 520 and 468 (i.e. LCM of the two i.e. 4680) be x
So,
x
+
17
=
4680
x
=
4663
x
–
17
=
4680
520
2
2
2
1
3
2
6
5
1
13
1
0
3
6
0
468
2
2
2
1
1
3
3
3
1
13
1
7
3
3
4
=
23
×
5
×
13
=
22
×
32
×
13
9
5