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# Let x and y be positive integers such that 7x^5 = 11y^13. The minimum

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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum  [#permalink]

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20 Mar 2019, 23:26
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32% (02:34) correct 68% (02:40) wrong based on 22 sessions

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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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Re: Let x and y be positive integers such that 7x^5 = 11y^13. The minimum  [#permalink]

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21 Mar 2019, 01:45
Bunuel wrote:
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

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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Re: Let x and y be positive integers such that 7x^5 = 11y^13. The minimum  [#permalink]

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21 Mar 2019, 04:40
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

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Let x and y be positive integers such that 7x^5 = 11y^13. The minimum  [#permalink]

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23 Mar 2019, 23:13
3
Archit3110 wrote:
Bunuel wrote:
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

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

Archit3110
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Re: Let x and y be positive integers such that 7x^5 = 11y^13. The minimum  [#permalink]

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24 Mar 2019, 18:20
Bunuel wrote:
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

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.

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Re: Let x and y be positive integers such that 7x^5 = 11y^13. The minimum  [#permalink]

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06 Apr 2019, 08:20
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?

GMATinsight wrote:
Archit3110 wrote:
Bunuel wrote:
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

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

Archit3110
Re: Let x and y be positive integers such that 7x^5 = 11y^13. The minimum   [#permalink] 06 Apr 2019, 08:20
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