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Math Expert V
Joined: 02 Sep 2009
Posts: 60515
On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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45 00:00

Difficulty:   85% (hard)

Question Stats: 58% (02:45) correct 43% (02:47) wrong based on 240 sessions

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On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders will be:[/b]

(A) 3, 3, 2
(B) 3, 4, 2
(C) 5, 4, 3
(D) 5, 5, 2
(E) 6, 4, 3

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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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18
5
Bunuel wrote:
On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders will be:[/b]

(A) 3, 3, 2
(B) 3, 4, 2
(C) 5, 4, 3
(D) 5, 5, 2
(E) 6, 4, 3

Here, the concept is of successive division
i.e. the no is first divided by 5 and it leaves remainder 2 and quotient is let x,

therefore we have $$number = 5x + 2$$ ---------------------- (1)

and then the quotient x is divided by 7 and the remainder is 3

so, we have x in the form of $$x = 7y + 3$$

and then the quotient y is divided by 8 and the remainder is 4

so, we have x in the form of $$y = 8z + 4$$

putting this value of x and y in (1) above, we get

$$number = 5(7(8z +4) + 3) +2$$
$$=> number = 5(56 z +31) +2$$
$$=> number = 280 z + 157$$

when this number will be divided by 8, we will get remainder = 5 and quotient = 35 z + 19

when this quotient will be divided by 7, we will get remainder = 5 and quotient = 5 z + 2

when this quotient will be divided by 5, we will get remainder = 2

Kudos, if you like the explanation
##### General Discussion
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On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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5
4
The number which we divided by 8 will be quotient from division by 7 = 8x + 4
The number which we divided by 7 will be quotient from division by 5 = 7(8x+4) + 3 = 56x +31
The original number which we divided by 5 =5(56x +31) + 2
= 280x +155 +2
= 280x + 157

When this number is divided by 8 = ( 280x+ 157)/8 = 35x + 19 and a remainder of 5
On dividing by 7 , ( 35x + 19) / 7 = 5x + 2 and a remainder of 5
On Dividing by 5 , (5x+2)/5 = x and a remainder of 2

Therefore the remainders will be 5,5 and 2

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On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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1
2
Bunuel wrote:
On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders will be:[/b]

(A) 3, 3, 2
(B) 3, 4, 2
(C) 5, 4, 3
(D) 5, 5, 2
(E) 6, 4, 3

On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively

Third divisor = 8
Third remainder = 4
i.e. Number of this type = 8+4 = 12

Second divisor = 7
Quotient then = 12
Second remainder = 3
i.e. Number of this type = 12*7+3 = 87

First divisor = 5
Quotient then = 87
Second remainder = 2
i.e. Number of this type = 87*5+2 = 437

Now, successively divide by 8, 7 and 5

837 divided by 8 leaves Quotient 54 and remainder=5

54 divided by 7 leaves Quotient 7 and remainder=5

7 divided by 5 leaves Quotient 1 and remainder=2

Successive Remainders (5, 5, 2)

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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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5
2
Bunuel wrote:
On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders will be:[/b]

(A) 3, 3, 2
(B) 3, 4, 2
(C) 5, 4, 3
(D) 5, 5, 2
(E) 6, 4, 3

Cross Multiplication Formula

5 7 8
2 3 4

(4*7)+3=31
(31*5)+2 = 157

Number is 157

Successive division of 157 by 8,7 and 5 will give remainder 5,5 and 2.

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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders will be:

(A) 3, 3, 2
(B) 3, 4, 2
(C) 5, 4, 3
(D) 5, 5, 2
(E) 6, 4, 3

Let the original number be X. Then by the successive dividing we have followings :
X=5A+2
A=7B+3
B=8C+4.

So we have A=7*(8C+4)+3=7*8C+31, and X=5*(7*8C + 31)+2 = 5*7*8C + 157.

Now by dividing X by 8, 7, 5 successively we have followings :
X=8*(5*7C+19)+5
5*7C+19=7*(5C+2) + 5
5C+2=5C+2.

The remainders are, therefore, 5, 5, 2.

OR

As dividing X by 8, 7, 5, we can ignore 5*7*8C(since 5*7*8C is a common multiple of 8, 7, 5, 5*7*8C doesn’t affect the remainders ).
So we can consider only 157.
157=8*19 + 5
19=7*2 + 5
2=5*0+2.
The remainders are 5, 5, 2.
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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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this is the first time when I see such a question...
my approach:
X is the number
X=5Q+2
5Q+2 = 7Z +3
7Z+3 = 8M + 4

so X=8M+4

where did I go wrong?
I believe the question is not worded properly: is the quotient only divided or together with the reminder?
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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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mvictor, according to your approach you are dividing the same no. by 8,7 and 5 not successively.i.e.
x=5q+2,
5q+2=7z+3=x=8m+4
for dividing successively we need to divide quotient *multiple of the previous no. as explained by mathrevolution and skywalker18.
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On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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when number is divided by 8 remainder is 4, so number is = 4, which is Q quotient
when number which is divided by 7 and leaver a remainder 3 would be 7*4 +3 = 31 , which quotient for the number divided by 5
when number which is divided by 5 and leaver a remainder 2 would be 5*31 +2 = 157

Now 157 / 8 remainder is 5 and Q quotient is 19 , next is 19/7 remainder is 5
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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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1
1
1
For a discussion on the concept of successive division and this question, check:

http://www.veritasprep.com/blog/2012/10 ... -division/
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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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1
Bunuel wrote:
On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders will be:[/b]

(A) 3, 3, 2
(B) 3, 4, 2
(C) 5, 4, 3
(D) 5, 5, 2
(E) 6, 4, 3

In successive division, the quotient will become the next dividend.

Suppose number is N

Then N = 5x+2; Next dividend x.

x = 7y+3; Next dividend y

y = 8z+4

so N = 5* {7*(8z+4) +3} +2 = 5 * (56z+31) +2 = 280 z +157

When N is divided by 8, the quotient is: 8*(35z+19) + 5

First Remainder is 5.

So options A, B E are out.

Quotient is 35z+19; This becomes next dividend.

Divide it by 7: 7 *(5z+2) + 5

Remainder is 5. --> So option C is out

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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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Bunuel what is the significance of the line "When the order of division is reversed" in the question?
Math Expert V
Joined: 02 Sep 2009
Posts: 60515
Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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chandan1988 wrote:
Bunuel what is the significance of the line "When the order of division is reversed" in the question?

First we divide x by 5, 7 and 8 successively, and getting the remainders of 2, 3 and 4 respectively.
Then we are dividing x by 8, 7, and 5 successively(so we are dividing by the same divisors but in reverse order), and getting the remainders of 8, 7 and 5, respectively.

For more on successive dividing check Successive Division post.

Hope it helps.
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Re: On dividing a certain number by 5, 7 and 8 successively, the remainder  [#permalink]

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