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Division Problem

Expert replies
by Viren1808 » Fri Feb 10, 2012 2:54 am
A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided by 17?

1. 8
2. 9
3. 10
4. 11

Kindly provide a detailed solution.
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Source: — Problem Solving |

by LalaB » Fri Feb 10, 2012 4:18 am
X/238= m +79

x/17=?


X/238=m +79
let m=1

then x=238*1+79=317

317=18*17+11

answ is E
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by Anurag@Gurome » Fri Feb 10, 2012 4:28 am
Viren1808 wrote:A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided by 17?

1. 8
2. 9
3. 10
4. 11

Kindly provide a detailed solution.
238 + 79 = 317
317/17 gives a quotient of 18.
17 * 18 = 306 and 317 - 306 = 11

The correct answer is [spoiler](4)[/spoiler].
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by Viren1808 » Fri Feb 10, 2012 8:21 am
LalaB wrote:
317=18*17+11

answ is E
Can you please explain me this last step? How would I land to this "18*17+11"?
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by GMATGuruNY » Fri Feb 10, 2012 10:37 am
Viren1808 wrote:A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided by 17?

1. 8
2. 9
3. 10
4. 11

Kindly provide a detailed solution.
The smallest integer that will yield a remainder of x is THE REMAINDER ITSELF.

Thus, in the problem at hand, the remainder -- 79 -- is smallest possible value of the unknown number:
79/238 = 0 R79.

Dividing by 17, we get:
79/17 = 4 R11.

The correct answer is 4.
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by [email protected] » Sun Feb 12, 2012 4:51 pm
A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided by 17?

1. 8
2. 9
3. 10
4. 11


Yes the correct answer is (4)... this is how i did it.....

STEP 1: FACTORIZE THE NUMBER 238 AND FIND THE FACTORS OF THE NUMBER

the numbers are 2, 7 and 17...


STEP 2: NOW YOU KNOW THAT THE NUMBER IS A MULTIPLE OF 238, SO TAKE ANY NUMBER GREATER THAN OR

EQUAL TO 238

AND ITS MULTIPLE FOR EG: 238, (238 * 2) ETC.. AND THEN ADD THE REMAINDER TO THAT NUMBER


STEP 3: DIVIDE THAT NUMBER BY 17 AND WHATEVER IS THE REMAINDER, IS THE FINAL ANSWER...




HOPE THIS HELPS....
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