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UPSC 2014 CDS II Math Paper
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© examsnet.com
Question : 24
Total: 100
How many pairs of X and Y are possible in the number 763X4Y2, if the number is divisible by 9?
[2014 CDS-II]
8
9
10
11
Validate
Solution:
The number 763X4Y2 is divisible by 9 if
7
+
6
+
3
+
X
+
4
+
Y
+
2
=
X
+
Y
+
22
is divisible by 9.
So,
X
+
Y
=
5
or
X
+
Y
=
14
So, the possible pair of X and Y are (0,5),(1,4),(2,3),(3,2),(4,1),(5,0),(5,9),(6,8),(7,7),(8,6) and (9,5).
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