For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What is the value of (n - m) if m and n are four-digit numbers for which *m* = (3^r)(5^s)(7^t)(11^u) and *n* = (25)(*m*)?
a) 2000
b) 200
c) 25
d) 20
e) 2
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The answer is b.
When you multiply *m* by 25, it's the same thing as multiplying it by 5^2. 5^2 would fall into the second part of the term, 5^b and corresponds with the hundreds digit of the number m and n. If this is the only difference between the two numbers, then m-n would end up being 200. b.
When you multiply *m* by 25, it's the same thing as multiplying it by 5^2. 5^2 would fall into the second part of the term, 5^b and corresponds with the hundreds digit of the number m and n. If this is the only difference between the two numbers, then m-n would end up being 200. b.
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m=1000r+100s+10t+u
*n*=5^2*(3^r)(5^s)(7^t)(11^u)=(3^r)(5^(s+2))(7^t)(11^u)
hence n=1000r+100(s+2)+10t+u
n-m=200
ans option B
*n*=5^2*(3^r)(5^s)(7^t)(11^u)=(3^r)(5^(s+2))(7^t)(11^u)
hence n=1000r+100(s+2)+10t+u
n-m=200
ans option B
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How did you know that 3^r for example corresponds to the thousands digit? also, would it be possible to elaborate on the breakdown a little more?
Could you also solve by plugging in numbers?
Could you also solve by plugging in numbers?
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In the question it is given thatpkw209 wrote:also, how did you jump from 5^s+2 to 100(s+2)?
*abcd*= (3^a)(5^b)(7^c)(11^d)
now for abcd ,a is the digit at thousandth place and so on...
if we map the same logic for *m* = (3^r)(5^s)(7^t)(11^u) i.e let m=abcd then
*m*=(3^a)(5^b)(7^c)(11^d)=(3^r)(5^s)(7^t)(11^u)..shows a=r or r is the digit at thousands place
as per number picking...even if we pickup numbers finally we have to get each digits of the number
ex let m=1000..then *m*=3
*n*=75=3^1*5^2..so n=1200
n-m=200
"If you don't know where you are going, any road will get you there."
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That's the explanation I was looking for. Extremely helpful.In the question it is given that
*abcd*= (3^a)(5^b)(7^c)(11^d)
now for abcd ,a is the digit at thousandth place and so on...
if we map the same logic for *m* = (3^r)(5^s)(7^t)(11^u) i.e let m=abcd then
*m*=(3^a)(5^b)(7^c)(11^d)=(3^r)(5^s)(7^t)(11^u)..shows a=r or r is the digit at thousands place
as per number picking...even if we pickup numbers finally we have to get each digits of the number
ex let m=1000..then *m*=3
*n*=75=3^1*5^2..so n=1200
n-m=200
Thanks!
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In here, m is the four-digit number rstu, and n is another four-digit number such thatpkw209 wrote:For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What is the value of (n - m) if m and n are four-digit numbers for which *m* = (3^r)(5^s)(7^t)(11^u) and *n* = (25)(*m*)?
a) 2000
b) 200
c) 25
d) 20
e) 2
*n* = (3^r){5^(s + 2)}(7^t)(11^u), or n's four digits are r(s + 2)tu, s is such that s + 2 is still a digit, now imagine the subtraction of rstu from r(s + 2)tu, all places find a 0 EXCEPT the hundreds' place, which will be 2, and hence the difference n - m will be [spoiler]200.
B[/spoiler]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com