How many positive integers divide 35^12 but not 35^11?

Question

A. 20
B. 22
C. 25
D. 23
E. 30

Bunuel · Accepted Answer

35^12 will obviously have all the factors of 35^11 and some more. 35^{12}=5^{12}*7^{12} --> # of factors is (12+1)(12+1)=169 . 35^{11}=5^{11}*7^{11} --> # of factors is (11+1)(11+1)=144 . The difference = 169 - 144 = 25. Answer: C. THEORY: Finding the Number of Factors of an Integer First make prime factorization of an integer n=a^p*b^q*c^r , where a , b , and c are prime factors of n and p , q , and r are their powers. The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1) . NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2 Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors. For more check: math-number-theory-88376.html Questions to practice: how-many-odd-positive-divisors-does-540-have-106082.html how-many-factors-does-36-2-have-126422.html how-many-different-positive-integers-are-factor-of-130628.html how-many-distinct-positive-factors-does-30-030-have-144326.html m02-72467.html Hope this helps.

sgangs · Answer

As usual Brunnel has a good solution ready.
However, I solved it in a different way.
I'm metioning it here as this method will help identify those "uncommon" factors.

35^11 is not divisible by 35^12.
Moreover, 35^11 has 7^11*5^11 as its factor, while 35^12 has 7^12*5^12 as its factor. (Highest powers of 7 &5 are 11 & 12 respectively)

Hence the remaining uncommon factors are 
7^12*5^0, 7^12*5^1,....,7^12*5^11 (remember 7^12,5^12 has already been considered as 35^12) , i.e. 12
and
5^12*7^0,5^12*7^1,...,5^12*7^11 (remember 7^12,5^12 has already been considered as 35^12) , i.e. 12

Hence total no of "uncommon" factors = 12+12+1=25

nphatak · Answer

The trick is to understand that they are looking for the number of factors for each of them . Factorize them and you will see that..

35^12 = 7^12 *5^12. The number of factors would be (12+1)(12+1) = 169
35^11 = 7^11 *5^11. The number of factors would be (11+1)(11+1) = 144

The difference is 25.

Another problem I though about. Difference in the sum of the positive integers that divide 35^2 and 35^3. Note the powers are different to make the calculations possible. 
If you read the GMATclub math book, you will see the formula to calculate the sum of the of factors of an integer is, if N= (a^m)(b^n)
Sum = (a^(m+1)-1)(b^(n+1)-1)/(a-1)(b-1)

You can factorize the calculations and finally get to workable numbers. I am almost certain that something like this will not come on GMAT.

JeffTargetTestPrep · Answer

Since all integers that divide 35^11 also divide 35^12, we need to determine how many more factors are in 35^12 than are in 35^11.

We can use the rule in which we break our bases to prime factors, add 1 to the exponent of each unique prime and then multiply those values together.

35^11 = 7^11 x 5^11

So, 35^11 has (11 + 1)(11 + 1) = 12 x 12 = 144 factors.

35^12 = 7^12 x 5^12

So, 35^12 has (12 + 1)(12 + 1) = 13 x 13 = 169 factors.

So, the number of integers that divide 35^12 but not 35^11 is 169 - 144 = 25.

Answer: C

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How many positive integers divide 35^12 but not 35^11?

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How many positive integers divide 35^12 but not 35^11?

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