MBA Exam

The HCF of three natural numbers x, y and z is 13. If the sum of x, y and z is 117, then how many ordered triplets (x, y, z) exist?

1) 27
2) 28
3) 54
4) 55

Read Solution (Total 1)

28

x+y+z= 13*(a+b+c)=117
a+b+c=9
Each of a,b,c >1.
a'+b'+c'=6 where a'=a-1, b'=b-1, c'=c-1

so take triplets for a',b',c'
no of triplets are (6+3-1)C(3-1)=8C2=28
12 years agoHelpfull: Yes(1) No(1)

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