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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 20 Mar 2019, 23:26
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Let x and y be positive integers such that \(7x^5 = 11y^{13}\). The minimum possible value of x has a prime factorization \(a^cb^d\). What is \(a + b + c + d\)?


(A) 30
(B) 31
(C) 32
(D) 33
(E) 34
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B
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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 23 Mar 2019, 23:13
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Archit3110
Bunuel
Let x and y be positive integers such that \(7x^5 = 11y^{13}\). The minimum possible value of x has a prime factorization \(a^cb^d\). What is \(a + b + c + d\)?


(A) 30
(B) 31
(C) 32
(D) 33
(E) 34

not able to solve question
GMATinsight sir please advise

given info we know factors of x = c+d+2
for what value of x & y will \(7x^5 = 11y^{13}\) is tricky to deduce..
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for \(7x^5 = 11y^{13}\) to be true for c and y to be integer

x must be a multiple of 11 for sure and y must be a multiple of 7

But since the powers need to be balanced so x must also be a multiple of 7 and y must be a multiple of 11 too

Let, \(x = 7^p*11^q\)
and \(y = 7^r*11^s\)

Now we have \(7x^5 = 11y^{13}\) as

\(7*7^{5p}*11^{5q} = 11*7^{13r}*11^{13s}\)

\(7^{5p+1}*11^{5q} = 7^{13r}*11^{13s+1}\)

now, 5p+1 = 13r . and 5q = 13s+1

p=5, r = 2 and q=8, s=3 are the first positive integer solutions of these two equations

i.e. \(x_{min} = 7^5*11^8\)

i.e. required sum = 7+5+11+8 = 31

Answer: Option B

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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 21 Mar 2019, 01:45
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Bunuel
Let x and y be positive integers such that \(7x^5 = 11y^{13}\). The minimum possible value of x has a prime factorization \(a^cb^d\). What is \(a + b + c + d\)?


(A) 30
(B) 31
(C) 32
(D) 33
(E) 34
Show more

not able to solve question
GMATinsight sir please advise

given info we know factors of x = c+d+2
for what value of x & y will \(7x^5 = 11y^{13}\) is tricky to deduce..
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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 21 Mar 2019, 04:40
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x = 11^n*7^m
y = 11^p*7^q

The equation translates in to:

11^5n*7^(5m+1)=11^(13p+1)*7^13q

so,
5m +1 =13q
and 5n = 13p+1
m, n, q and p are all integers

If m= 1 q = 6/13 X
m=2, q =11/13 X
m=3, q= 16/13 X
m=4, q=21/13 X
m = 5, q= 26/13= 2

If p = 1, n = 14/5
p=2 , n= 27/5
p=3, n = 40/5=8

So x =11^8*7^5

11+8+7+5 = 31 Answer B)
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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 24 Mar 2019, 18:20
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Bunuel
Let x and y be positive integers such that \(7x^5 = 11y^{13}\). The minimum possible value of x has a prime factorization \(a^cb^d\). What is \(a + b + c + d\)?


(A) 30
(B) 31
(C) 32
(D) 33
(E) 34
Show more

Since 7 is not divisible by 11, x must have a factor of 11, and likewise, since 11 is not divisible by 7, y must have a factor of 7. However, if y has a factor of 7, then the right hand side of the equation will have (at least) 13 factors of 7. Therefore, x must have (at least) a factor of 7. Similarly, if x has a factor of 11, then the left hand side of the equation will have (at least) 5 factors of 11. Therefore, y must have (at least) a factor of 11.

Therefore, we can let x = 11^c * 7^d (in order words, we are letting a = 11 and b = 7) and y = 11^s * 7^t. Now let’s substitute them into the equation 7x^5 = 11y^13:

7(11^c * 7^d)^5 = 11(11^s * 7^t)^13

11^(5c) * 7^(5d + 1) = 11^(13s + 1) * 7^(13t)

So we must have:

5c = 13s + 1 and 5d + 1 = 13t

Since we want the minimum possible value of x, we want positive integers c and d to be as small as possible, and hence positive integers s and t to be a small as possible. We see that the smallest positive integer value for s is 3, and in that case c will be 8, and the smallest positive integer value for t is 2, and in that case d will be 5.

Therefore, x = 11^8 * 7^5 and 11 + 7 + 8 + 5 = 31.

Answer: B
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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 06 Apr 2019, 08:20
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Please can you expantiate on on this statement " but since the powers need to be balanced so x must also be a multiple of 7 and y must be a multiple of 11 too"...Why do the powers need to be balanced?

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Archit3110
Bunuel
Let x and y be positive integers such that \(7x^5 = 11y^{13}\). The minimum possible value of x has a prime factorization \(a^cb^d\). What is \(a + b + c + d\)?


(A) 30
(B) 31
(C) 32
(D) 33
(E) 34

not able to solve question
GMATinsight sir please advise

given info we know factors of x = c+d+2
for what value of x & y will \(7x^5 = 11y^{13}\) is tricky to deduce..

for \(7x^5 = 11y^{13}\) to be true for c and y to be integer

x must be a multiple of 11 for sure and y must be a multiple of 7

But since the powers need to be balanced so x must also be a multiple of 7 and y must be a multiple of 11 too

Let, \(x = 7^p*11^q\)
and \(y = 7^r*11^s\)

Now we have \(7x^5 = 11y^{13}\) as

\(7*7^{5p}*11^{5q} = 11*7^{13r}*11^{13s}\)

\(7^{5p+1}*11^{5q} = 7^{13r}*11^{13s+1}\)

now, 5p+1 = 13r . and 5q = 13s+1

p=5, r = 2 and q=8, s=3 are the first positive integer solutions of these two equations

i.e. \(x_{min} = 7^5*11^8\)

i.e. required sum = 7+5+11+8 = 31

Answer: Option B

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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 25 Mar 2021, 19:02
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Hi,

7x^5=11y^13
as x= a^c*b^d , x^5=a^5c*b^5d
and because 11 is a prime factor of x^5 and x contains two primes, a and b must be 7 and 11

x^5 becomes 7^5c*11^5d
In the same way, y^13 becomes 7^13w*11^13z

Rewriting the equation according to these elements gives us:
7 * 7^5c * 11^5d= 11 * 7^13w * 11^13z
11/7=(7^5c * 11^5d)/(7^13w * 11^13z)

We get (11^5d)/(11^13z) =11
5d-13z=1
5d=13z+1
d=(13z+1)/5 here because we want the minimal value for x and therefore for c and d, we will take the minimal value for z that makes the equation divisible by 5. In our case it is z=3.
d=(13*3+1)/5 = 40/5 = 8

Same process for finding c:
7^5c/7^13w= 1/7 or 7^-1
5c-13w=-1
5c=13w-1
c=(13w-1)/5 We get z=2
c=(13*2-1)/5= 25/5=5

Finally! We have a=7 b=11 c=5 and d=8
7+11+5+8=31

Answer B
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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum 01 Nov 2025, 08:47
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