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Re: What is the best way to tackle this kind of DS problems [#permalink]
07 Feb 2012, 06:31

Expert's post

1

This post was BOOKMARKED

smodak wrote:

If i and d are integers, what is the value of i? (1) The remainder when i is divided by (d+2) is the same as when i is divided by d (2) The quotient when i is divided by (d+2) is d

What is the best way to tackle this kind of DS problems?

It can be done algebraically but picking numbers would probably be faster/easier.

If i and d are integers, what is the value of i?

(1) The remainder when i is divided by (d+2) is the same as when i is divided by d --> let the remainder be 0 to simplify the case. So, we have that i is divisible by both d and d+2 --> if d=1 then d+2=3 and i can be ANY multiple of 3. Not sufficient.

(2) The quotient when i is divided by (d+2) is d --> let the remainder be 0 to simplify the case. Now, if d=2 then d+2=4, so i=8 (8/4=2: i=8 divided by d+2=4 yields the quotient of d=2) but if d=3 then d+2=5 and i=15 (15/5=3). Not sufficient.

(1)+(2) Notice that two values of i from (2) works for (1) as well: 8 is divisible d=2 and d+2=4 and 15 is divisible by d=3 and d+2=5.

Re: What is the best way to tackle this kind of DS problems [#permalink]
13 May 2014, 15:12

Bunuel wrote:

smodak wrote:

If i and d are integers, what is the value of i? (1) The remainder when i is divided by (d+2) is the same as when i is divided by d (2) The quotient when i is divided by (d+2) is d

What is the best way to tackle this kind of DS problems?

It can be done algebraically but picking numbers would probably be faster/easier.

If i and d are integers, what is the value of i?

(1) The remainder when i is divided by (d+2) is the same as when i is divided by d --> let the remainder be 0 to simplify the case. So, we have that i is divisible by both d and d+2 --> if d=1 then d+2=3 and i can be ANY multiple of 3. Not sufficient.

(2) The quotient when i is divided by (d+2) is d --> let the remainder be 0 to simplify the case. Now, if d=2 then d+2=4, so i=8 (8/4=2: i=8 divided by d+2=4 yields the quotient of d=2) but if d=3 then d+2=5 and i=15 (15/5=3). Not sufficient.

(1)+(2) Notice that two values of i from (2) works for (1) as well: 8 is divisible d=2 and d+2=4 and 15 is divisible by d=3 and d+2=5.

Answer: E.

Hope it's clear.

What's the trick in this question? Is it only the fact that 'd' and 'd+2' as denominators can be larger than 'i' and therefore, 'i' could take any value as long as it is smaller than the denominator as well as being a multiple of both 'd' and 'd+2' ? (Has to be a multiple otherwise remainder can't be zero)

Re: What is the best way to tackle this kind of DS problems [#permalink]
30 May 2014, 05:07

Bunuel wrote:

smodak wrote:

If i and d are integers, what is the value of i? (1) The remainder when i is divided by (d+2) is the same as when i is divided by d (2) The quotient when i is divided by (d+2) is d

What is the best way to tackle this kind of DS problems?

It can be done algebraically but picking numbers would probably be faster/easier.

If i and d are integers, what is the value of i?

(1) The remainder when i is divided by (d+2) is the same as when i is divided by d --> let the remainder be 0 to simplify the case. So, we have that i is divisible by both d and d+2 --> if d=1 then d+2=3 and i can be ANY multiple of 3. Not sufficient.

(2) The quotient when i is divided by (d+2) is d --> let the remainder be 0 to simplify the case. Now, if d=2 then d+2=4, so i=8 (8/4=2: i=8 divided by d+2=4 yields the quotient of d=2) but if d=3 then d+2=5 and i=15 (15/5=3). Not sufficient.

(1)+(2) Notice that two values of i from (2) works for (1) as well: 8 is divisible d=2 and d+2=4 and 15 is divisible by d=3 and d+2=5.

Answer: E.

Hope it's clear.

Can this somehow be done algebraically or conceptually

I had that first i/ (d+2) and i/d, yield the same remainder, but if both d and d+2, are larger than i, then 'i' could just take any value as the remainder.

Clearly insufficient

Statement 2 we have that i = (d)(d+2) + r

Now, if we replace in first term, we have that (d)(d+2) + r / (d+2), gives d as quotient but still we are left with r as a remainder of d+2. while we also know that i/d gives the same remainder. Here we learn that 'i' must be greater than d+2 since if gives quotient d. However, we still have no clue as to what remainder the division can yield

Both together, i>d, i = d(d+2) + r, and 'r' here is equal to the remainder of i/d = d(d+2) / d.

So r/d = r/d+2, but still no information on the remainder

Re: If i and d are integers, what is the value of i? [#permalink]
17 Jun 2014, 21:05

Expert's post

smodak wrote:

If i and d are integers, what is the value of i?

(1) The remainder when i is divided by (d+2) is the same as when i is divided by d (2) The quotient when i is divided by (d+2) is d

What is the best way to tackle this kind of DS problems?

Question: What is the value of i?

How do you express stmnt 1 in an equation? Stmnt 1: The remainder when i is divided by (d+2) is the same as when i is divided by d.

Say when i is divided by d or d + 2, the remainder we obtain is r. Does this mean that if we subtract r from i, whatever is leftover will be divisible by d as well as (d+2)? So assuming that d and d+2 do not have any common factors (even if they do have common factors other than 1, they can only have 2 as a common factor), we can put it down as

i - r = d(d + 2)k i = d(d + 2)k + r Now for different values of d, k and r, values of i will be different.

Stmnt 2: The quotient when i is divided by (d+2) is d This tells us that i = d(d +2) + r Now for different values of d and r, values of i will be different.

Using both, we know that k is 1. But for different values of d and r, we can still have different values of i. Not sufficient.

Re: If i and d are integers, what is the value of i? [#permalink]
01 Jul 2015, 05:37

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