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If A,BCD is a four-digit positive integer such that A,BCD is equal to

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If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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New post 30 Aug 2015, 10:52
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Question Stats:

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If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

I. AD
II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III


Kudos for a correct solution.

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Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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New post 01 Sep 2015, 23:37
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Bunuel wrote:
If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

I. AD
II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III


Kudos for a correct solution.


Ideally, ABC * AB = ABCD should make you think of 1s and 0s. The digits needn't be unique so the first thing that comes to mind is 100*10 = 1000. But since CA and CD are given as two digit numbers, next you should think of 101*10 = 1010 or any such number. This satisfies all conditions so we can analyse using this example.

101 * 10 = 1010

AD = 10 - a factor of 1010
CD = 10 - a factor of 1010
CA = 11 - not a factor of 1010

Now the only thing left to figure is whether it is possible that C1 is a factor of 10C0 for some value of C.

10C0 = 1000 + C0 = 999 + C1
C1 is divisible by C1 but we don't know whether 999 is divisible by C1.
999 = 3^3*37
It has no two digit factor ending in 1 so it will not be divisible by C1.

Answer (C)



In case, 1s and 0s don't come to mind, here is how you could arrive at 10C0:

A bit of visualisation will help:

........A B C
........x A B
__________________
........M N P
.....Q R S x
__________________
.....A B C D


Note the leftmost digits: A*A gives you Q and you get A in the final product.
1*1 = 1 (possible value of Q and A)
2*2 = 4 (even if there were a carry over from previous calculation of M + R, Q+1 = 5, not A(2)). Hence A = 2 is not possible.
3*3 = 9 (even if there were a carry over from previous calculation, Q+1 would be 10 and the final product would be a 5 digit number.)

A cannot be any larger value because then the product would be a 5 digit number. Hence A must be 1 and there must be NO carryover from the previous calculation.

........1 B C
........x 1 B
__________________
........B N P
.....1 B C x
__________________
.....1 B C D

Now note that B+B = B. This is only possible if B = 0 or B = 9 (with a carryover from previous addition). But 9 + 9 + 1 = 19 and that would give a carryover to the next addition which we don't want. So B must be 0

So you get 10C * 10 = 10C0

This is true for all values of C from 1 to 9. (C cannot be 0 because CD is a two digit number)
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If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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New post 31 Aug 2015, 03:20
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1
Bunuel wrote:
If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

I. AD
II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III


Kudos for a correct solution.



IMO : C

AB=10*A+B
ABC=100*A+10*B+C
ABCD=1000*A+100*B+10*C+D
AB*ABC=1000A^2 +100AB+10AC+100AB+10B^2+BC
=ABCD=1000*A+100*B+10*C+D
SO,A^2=A SO A=1
100AB=100B
10AC=10C
SO D=100AB+10B^2+BC
This is possible only if B=0 SO,D=0
Hence A=1,B=0,D=0 C can be any digit.
From options AD=10 which is always a factor of ABCD since D=0
CD may or may not be a factor ( if C = 5 then yes, if 6 then no)

CA cannot be a factor of ABCD. (If c = 5 then 51 not a factor of 1050)

So only CA cant be the factor.CD may be factor.
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Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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New post 01 Sep 2015, 19:46
ABCD = ABC * AB
1000A + 100B + 10C + D = (100A + 10 B + C)* (10A + B)
= 1000 A^2 + 100 (2AB) + 10( B^2 +AC) + BC
=> 1000 (A^2 - A) + 100 (2AB - B) + 10 ( AC + B^2 - B) + (BC - D) = 0
Given, A >0 ans as B, C , D are single digit integers which means(BC- D) will be atmost two digit number, for LHS to be zero, as the first two co-efficients are positive they need to be zero ( A^2 >=A & 2AB > B for all values).
2AB = B => A = 1/2 or B = 0.. As A is integer , B= 0
A^2 - A = 0 => A = 1

As D is product of B & C => D is 0

The number is of the form 10C0
Any value of C would be satisfy

AD (10 ) is factor of 10C0
CD (C0) if C is 3, 30 is not factor of 1030
CA (C1) if C is 1, 11 is not factor of 1010
Hence only AD is factor.

option D
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Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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New post 27 Jul 2017, 09:04
VeritasPrepKarishma wrote:
Bunuel wrote:
If A,BCD is a four-digit positive integer such that A,BCD is equal to the product of the two-digit number AB and the three digit number ABC, which of the following two-digit numbers cannot be a factor of A,BCD? (Note: the first digit of a number cannot be equal to 0.)

I. AD
II. CD
III. CA

(A) None
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III


Kudos for a correct solution.


Ideally, ABC * AB = ABCD should make you think of 1s and 0s. The digits needn't be unique so the first thing that comes to mind is 100*10 = 1000. But since CA and CD are given as two digit numbers, next you should think of 101*10 = 1010 or any such number. This satisfies all conditions so we can analyse using this example.

101 * 10 = 1010

AD = 10 - a factor of 1010
CD = 10 - a factor of 1010
CA = 11 - not a factor of 1010

Now the only thing left to figure is whether it is possible that C1 is a factor of 10C0 for some value of C.

10C0 = 1000 + C0 = 999 + C1
C1 is divisible by C1 but we don't know whether 999 is divisible by C1.
999 = 3^3*37
It has no two digit factor ending in 1 so it will not be divisible by C1.

Answer (C)



In case, 1s and 0s don't come to mind, here is how you could arrive at 10C0:

A bit of visualisation will help:

........A B C
........x A B
__________________
........M N P
.....Q R S x
__________________
.....A B C D


Note the leftmost digits: A*A gives you Q and you get A in the final product.
1*1 = 1 (possible value of Q and A)
2*2 = 4 (even if there were a carry over from previous calculation of M + R, Q+1 = 5, not A(2)). Hence A = 2 is not possible.
3*3 = 9 (even if there were a carry over from previous calculation, Q+1 would be 10 and the final product would be a 5 digit number.)

A cannot be any larger value because then the product would be a 5 digit number. Hence A must be 1 and there must be NO carryover from the previous calculation.

........1 B C
........x 1 B
__________________
........B N P
.....1 B C x
__________________
.....1 B C D

Now note that B+B = B. This is only possible if B = 0 or B = 9 (with a carryover from previous addition). But 9 + 9 + 1 = 19 and that would give a carryover to the next addition which we don't want. So B must be 0

So you get 10C * 10 = 10C0

This is true for all values of C from 1 to 9. (C cannot be 0 because CD is a two digit number)




if we put 107 * 10 = 1070 , in this case C is 7 and D is 0
then 1070 is not divisible by 70
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Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to  [#permalink]

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New post 27 Jul 2017, 22:11
optimusprime07 wrote:
if we put 107 * 10 = 1070 , in this case C is 7 and D is 0
then 1070 is not divisible by 70


Yes, but there exists a case in which it is. In the case of 101*10 = 1010, C = 1 and D = 0.
1010 is divisible by 10.
But the question asks: "which of the following two-digit numbers cannot be a factor of A,BCD?"
To qualify, CD should not be a factor of ABCD for all values of CD.
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Re: If A,BCD is a four-digit positive integer such that A,BCD is equal to   [#permalink] 27 Jul 2017, 22:11
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