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The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l , and m, respectively. If (2a)(3b)(5c) = 12(2k)(3l)(5m), what is the value of x – y?
  • a)
    21
  • b)
    200
  • c)
    210
  • d)
    300
  • e)
    310
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The three-digit positive integer x has the hundreds, tens, and units d...
First, let us simplify the exponential equation:

When the bases on both sides of an equation are equal and the bases are prime numbers, the exponents of the respective bases must also be equal: a = k + 2; b = l + 1; and c = m . Now recall that a , b , and c represent the hundreds, tens, and units digits of the three-digit integer x ; similarly, k , l , and m represent the hundreds, tens, and units digits of the three-digit integer y . 
Therefore, the hundreds digit of x is 2 greater than the hundreds digit of y ; the tens digit of x is 1 greater than the tens digit of y ; finally, the units digit of x is equal to the units digit of y . Using this information, we can set up our subtraction problem and find the value of (x – y ):
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Most Upvoted Answer
The three-digit positive integer x has the hundreds, tens, and units d...
We have that $12(2k)(3l)(5m) = (2a)(3b)(5c)$. Dividing both sides by $12$, we have $(2k)(3l)(5m) = (a)(b)(c)$. Since $x$ and $y$ are three-digit positive integers, we have $100 \leq x,y \leq 999$, so $1 \leq a,b,c,k,l,m \leq 9$, which means that $2k$, $3l$, and $5m$ are all factors of $(a)(b)(c)$.

Since $2k$ is a factor of $(a)(b)(c)$, we know that $k$ is a factor of $(a)(b)(c)/2$. Since $a$, $b$, and $c$ are single-digit positive integers, we know that $(a)(b)(c)/2$ is a multiple of $1/2 = 1$, $2$, $3$, $4$, or $6$. Therefore, $k$ is a factor of $1$, $2$, $3$, $4$, or $6$. Since $k$ is a single-digit positive integer, we must have $k = 1$, $k = 2$, $k = 3$, $k = 4$, or $k = 6$.

If $k = 1$, then $(2k)(3l)(5m) = 30lm$, and $(a)(b)(c) = 30lm$. Since $1 \leq l,m \leq 9$, we know that $30lm$ is a multiple of $30$, but it is not divisible by $100$, so $x \neq 30lm$.

If $k = 2$, then $(2k)(3l)(5m) = 60lm$, and $(a)(b)(c) = 60lm$. Since $1 \leq l,m \leq 9$, we know that $60lm$ is a multiple of $60$, but it is not divisible by $100$, so $x \neq 60lm$.

If $k = 3$, then $(2k)(3l)(5m) = 180lm$, and $(a)(b)(c) = 180lm$. Since $1 \leq l,m \leq 9$, we know that $180lm$ is a multiple of $180$, but it is not divisible by $100$, so $x \neq 180lm$.

If $k = 4$, then $(2k)(3l)(5m) = 120lm$, and $(a)(b)(c) = 120lm$. Since $1 \leq l,m \leq 9$, we know that $120lm$ is a multiple of $120$, but it is not divisible by $100$, so $x \neq 120lm$.

If $k = 6$, then $(2k)(3l)(5m) = 180lm$, and $(a)(b)(c) = 180lm$. Since $1 \leq l,m \leq 9$, we know that $180lm$ is a multiple of $
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The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. Thethree-digit positive integer y has the hundreds, tens, and units digits of k, l , and m, respectively. If(2a)(3b)(5c) = 12(2k)(3l)(5m), what is the value of x – y?a)21b)200c)210d)300e)310Correct answer is option 'C'. Can you explain this answer?
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