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Question

Suppose a2,a3,a4,a5,a6,a7 are integers such that
57=a22!+a33!+a44+a55!+a66!+a77!
where 0a<j for j=2,4,5,6,7. The sum a2+a3+a4+a5+a6+a7 is

A
8
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B
9
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C
10
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D
11
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Solution

The correct option is C 9

57=a22!+a33!+a44!+a55!+a66!+a77!57=12(a2+13(a3+14(a4+......(a77))))107=a2+13(a3+14(a4+......(a77)))1+37=a2+13(a3+14(a4+......(a77)))

So, as expression is less than 1

a2=137=13(a3+14(a4+......15(a5+.....a77)))1+27=a3+14(a4+......15(a5+16(a6+a77)))a3=1

Similarly

a4=1a7=2a5=0a6=4

So, Sum a71+1+1+4+2=9


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