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pablovaldesvega
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Just list out the ways of summing to 13 or 17- really not that bad:

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pablovaldesvega
How many different positive two-digit integers are there such that the sum of the two digits is a prime number greater than 11?

A) Four
B) Five
C) Six
D) Seven
E) Eight
Total Two digit numbes = 99-9 = 90

Two digit Prime Numbers = {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 79, 83, 89, 93, 97}

Sum of the two digits of any two-digit number can NOT exceed 9+9 = 18

so we are looking at only those two-digit numbers whose digit sum is {13, 17}

for 13: the two digit numbers may be {49, 94}4+9; {58, 85} 5+8; {67, 76} 6+7 i.e. 6 numbers
for 17 the two digit numbers can be 89 and 98 i.e. 2 numbers

so total favorable numbers = 6+2 = 8

Answer: Option E
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pablovaldesvega
How many different positive two-digit integers are there such that the sum of the two digits is a prime number greater than 11?

A) Four
B) Five
C) Six
D) Seven
E) Eight

Looks worse than it is. The constraint is that sum of digits should be a prime number > 11
So sum of digits could be 13 or 17 (not 19 or greater because 2 single digits cannot give the sum of 19 or greater).

17 is easy - only 1 way of getting a sum of 17. The digits should be {8, 9}. So we can make 2 numbers (89, 98)
13 - start with distributing the sum as equally as possible - {6, 7}, {5, 8}, {4, 9}. So we can make 6 numbers (same logic as above)

Hence total we can make 8 numbers.

Answer (E)
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