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For a positive integer n, P(n) is denoted as the product of

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For a positive integer n, P(n) is denoted as the product of

by Max@Math Revolution » Wed Sep 25, 2019 11:24 pm
[GMAT math practice question]

For a positive integer n, P(n) is denoted as the product of each digit of n. For example, P(29)=2*9=18, P(457)=4*5*7=140. a, b and c are 2-digit positive integers with P(a)P(b)P(c)=9. What is the maximum value of a+b+c?

A. 91
B. 102
C. 113
D. 124
E. 135
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by deloitte247 » Fri Sep 27, 2019 10:56 pm
P(a) * P(b) * P(c)=9
Factors of a= 1, 3, and 9
Therefore, the possible values of P(a), P(b) and P(c) could be;
P(a) = P(11) = 1*1 = 1
P(b) = P(11) = 1*1 = 1
P(c) = P(91) = 9*1 = 9
===> P(a) * P(b) * P(c) = 1*1*9 = 9
OR
P(a) = P(11) = 1*1 = 1
P(b) = P(31) = 3*1 = 3
P(c) = P(31) = 3*1 = 3
===> P(a) * P(b) * P(c) = 1*3*3 = 9
OR
P(a) = P(33) = 3*3 = 9
P(b) = P(11) = 1*1 = 1
P(c) = P(11) = 1*1 = 1
===> P(a) * P(b) * P(c) = 9*1*1 = 9
OR
P(a) = P(11) = 1*1 = 1
P(b) = P(19) = 1*9 = 9
P(c) = P(11) = 1*1 = 1
===> P(a) * P(b) * P(c) = 1*9*1 = 9

Out of all these possible values, the maximum value of (a+b+c) will occur when P(a)=P(11), P(b)=P(11), and P(c)=P(91), thereby making a=11; b=11 and c=91.
Therefore, (a+b+c) = 11 +11 +91 = 113

Option C is the correct answer.

Hope this helps? Thanks
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by Max@Math Revolution » Sun Sep 29, 2019 5:18 pm
=>

In order to have a maximum value of P(a)P(b)P(c), the possible cases of tuples (a,b,c) we should consider are (91,11,11) and (31,31,11).
For (91,11,11), we have 91 + 11 + 11 = 113.
For (31,31,11), we have 31 + 31 + 11 = 73.
Thus, the maximum of a + b + c is 113.

Therefore, C is the answer.
Answer: C
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