On dividing a certain number by 5, 7 and 8 successively, the remainder

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

Answer choice D

Kudos, if you like the explanation

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

Answer D
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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)

Answer: option D

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.

ANSWER IS D.

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.
The answer is (D).

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.

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?

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.

to find a number first start with division by 8
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
D is the answer.

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

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/10 ... -division/

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

D is the answer.

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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Hi VeritasKarishma, Bunuel!

I got this right! and I understand the set up and all of the logic perfectly, but it took me ~3 to 4 minutes to get.

Do you recommend anything to get this time down?

Thanks!!!

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

N = 5a + 2 = 5(7b + 3) + 2 = 5(7(8c + 4) + 3) + 2

To get first such number, put c = 0.
First such number = 157

157/8 gives quotient 19 and leaves remainder 5 (between options (C) and (D))
19/7 leaves remainder 5

Answer (D)

Shouldn't take more than 2 mins.

thanks!

I practiced it a few times and got it no problem!

Great post!

Solution:

We need to find a number that when it is divided by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively.

We have to go backward. That is, first we have to find a positive number that, when divided by 8, yields a remainder of 4. We see that the smallest such number is 4. Next, we need to find a number that, when divided by 7, yields a quotient of 4 and a remainder of 3. We see that number is 7 x 4 + 3 = 31. Finally, we need to find a number that, when divided by 5, yields a quotient of 31 and a remainder of 2. We see that number is 5 x 31 + 2 = 157.

Since we’ve found 157 using a “backward” process. We can check whether it actually works:

157/5 = 31 R 2

31/7 = 4 R 3

4/8 = 0 R 4

So 157 actually works. Now we can divide it by 8, 7, and 5, successively, and find the respective remainders:

157/8 = 19 R 5

19/7 = 2 R 5

2/5 = 0 R 2

Alternate Solution:

Let x be the number. Since x yields a remainder of 2 when divided by 5, we can write x = 5k + 2 for some positive integer k.

Since k yields a remainder of 3 when divided by 7, we can write k = 7s + 3 for some positive integer s.

Since s yields a remainder of 4 when divided by 8, we can write s = 8m + 4 for some positive integer m.

Let’s substitute s = 8m + 4 in k = 7s + 3:

k = 7(8m + 4) + 3

k = 56m + 28 + 3

k = 56m + 31

Let’s substitute k = 56m + 31 in x = 5k + 2:

x = 5(56m + 31) + 2

x = 280m + 155 + 2

x = 280m + 157

Now, let’s find the remainder when x is divided by 8. Let’s write:

x = 8 * 35m + 152 + 5

x = 8 * 35m + 8 * 19 + 5

x = 8(35m + 19) + 5

So, the remainder when x is divided by 8 is 5. We eliminate answer choices A, B and E.

Next, let’s divide the quotient from division by 8 (which is 35m + 19) by 7:

35m + 19 = 7* 5m + 14 + 5

35m + 19 = 7(5m + 2) + 5

So, the remainder when 35m + 19 is divided by 7 is also 5. We eliminate answer choice C and conclude that the correct answer is D.

Answer: D

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