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mSKR
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what Updated on: 13 Aug 2021, 07:00
Originally posted by mSKR on 13 Aug 2021, 03:16.
Last edited by mSKR on 13 Aug 2021, 07:00, edited 1 time in total.
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26% (02:38) correct 74% (02:34) wrong based on 161 sessions

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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 1-9). what is the value of C?

A. 1
B. 2
C. 7
D. 8
E. 9
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E
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what 13 Aug 2021, 08:10
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Questions such as this had better not appear in actual GMAT test. I can figure out A =2 and E = 8 because
(1) A can only be 1 or 2. Otherwise, ABCDE * 4 will result in five-digit number.
(2) It follows that E must be 4 and higher, and E * 4 should end in A.
4 * 4 = 16. 6 is not 1 or 2.
5 * 4 = 20. 0 is not 1 or 2.
6 * 4 = 24. 4 is not 1 or 2.
7 * 4 = 28. 8 is not 1 or 2
8 * 4 = 32. Found one possibility!
9 * 4 = 36. 6 is not 1 or 2.
So, we have A =2 and E = 8.

Beyond that, with a glance at the time already spent, I guess it is time to guess and move on. :-)

But, we can actually figure out what C is without knowing values for ABDE.

The key is that the same C is on the thousand unit. With the exception of 0 (which is not in the answer choices), a digit A and 4A do not share the same unit digit. What gives here? It is made possible thanks to the carry forward from "...DE" * 4.

The maximum carry is 3 when DE is >= 75 (300 / 4 = 75).

Now check five answer options:
A. 1. 1X4 = 4 and requires a carry of 7
B. 2 2X4 = 8 and requires a carry of 4
C. 7 7*4 = 8 and requires a carry of 4
D. 8 8 *4 = 2 and requires a carry of 6
E. 9 9 * 4 = 6 and requires a carry of 3

So, only E is possible.
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what 13 Aug 2021, 05:22
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Could someone kindly explain the process?
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what 13 Aug 2021, 07:23
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Could someone kindly explain the process?
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the answer => 21978 * 4= 87912


calculate A and E
start with A and E --> the values can be 2 and 8 respectively ( 1 and 4 would fail )


Then find B and D .
the equation would be 4*D +3 = B. but you don't want to add any carried forward to A in LHS . it means we need to restrict digit B to 1 or 2 ( if you multiple by 4 then no carry forward) .

D= 1 then B = 7 , if D = 2 then B = 11 , D= 3 then B 15, d= 4 then B 9 (19), D= 5 then B 23, D=6 then B 27, if d= 7 then B 31, if D=8 then B 35, if D=9 then B 39
notice last digit can be 1 or 2
only possiblilities (B,D) = 1,2 OR (1,7)
Our number can be:
21C28 *4 = 82C12--> this option can be rejected because when u multiple 21*4 then minimum value should be 84 but we can have only 82
21C78 * 4 = 87C12--> so far so good


FInally , need to calculate C:
4 * c+3 =3c ( 4*1 + carry forward should be equal to 7-i.e. RHS D)

we need final result from 31 to 39 as we need to keep 3 to carry forward .
Hence C has to be 9 : 4 * 9 + 3 = 39

Hence C = 9
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what 13 Aug 2021, 09:21
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mSKR
If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 1-9). what is the value of C?

A. 1
B. 2
C. 7
D. 8
E. 9
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ABCDE*4=EDCBA

A: Now, A is the unit's digit of 4*E, so A has to be EVEN.
Also, A*4+(carry over if any) is digit E, so A can be ONLY 1 or 2.
Since A is even, A is 2.

E: 2 is the unit's digit of 4*E, so E has to be 3 or 8.
Also, 2*4+(carry over if any) is digit E, so E cannot be 3. E=8.

2BCD8*4=8DCB2
Carryover to tens digit => 4*8=32, so 3.
B: Units digit of 4*D+3, which will be ODD, is B, so B can be 1, 3, 5, 7 or 9.
There is no carry over to ten thousands digit E in EDCBA, so 4*B+(carry over)=D.
Thus, B is 1.

D: Now 4*D+3 has units digit B or 1. Thus 4*D has a units digit of 8, giving possible values of D as 2 or 7.
Now, 4*B+(carry over)=D, so 4*1+(carry over)=D= 2 or 7.
D>4, so D=7 and carry over =3.

21C78*4=87C12

C: 4*C+3=(carry over 3)+C=30+C => 3C=27
Thus, C=9

E


There is no likelihood of such a question on GMAT. But use it to enhance your skills on number properties.
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what 13 Aug 2021, 19:12
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 1-9). what is the value of C?

A. 1
B. 2
C. 7
D. 8
E. 9
Show more


A shorter solution.

ABCDE*4 = EDCBA

So when you multiply 4*D on left side, the answer would be <40, so the carry over to 4*C could be 0, 1, 2 or 3.

Next when you multiply 4*C and add carry over, which could be 0, 1, 2 or 3, the units digit is again C.
That is, 4*C+(carry over 0-3)=xC

Let us use options now
A. 1……. 4*1+(carry over 0-3) will give units digit as 4, 5, 6 or 7. It will never be 1.
B. 2 ……. 4*2+(carry over 0-3) will give units digit as 8, 9, 0 or 1. It will never be 2.
C. 7 ……. 4*7+(carry over 0-3) will give units digit as 8, 9, 0 or 1. It will never be 7.
D. 8 ……. 4*8+(carry over 0-3) will give units digit as 2, 3, 4 or 5. It will never be 8.
E. 9 ……. 4*9+(carry over 0-3) will give units digit as 6, 7, 8 or 9. It can be 9….CORRECT


E
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what 15 Aug 2021, 01:56
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mSKR
If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 1-9). what is the value of C?

A. 1
B. 2
C. 7
D. 8
E. 9


A shorter solution.

ABCDE*4 = EDCBA

So when you multiply 4*D on left side, the answer would be <40, so the carry over to 4*C could be 0, 1, 2 or 3.

Next when you multiply 4*C and add carry over, which could be 0, 1, 2 or 3, the units digit is again C.
That is, 4*C+(carry over 0-3)=xC

Let us use options now
A. 1……. 4*1+(carry over 0-3) will give units digit as 4, 5, 6 or 7. It will never be 1.
B. 2 ……. 4*2+(carry over 0-3) will give units digit as 8, 9, 0 or 1. It will never be 2.
C. 7 ……. 4*7+(carry over 0-3) will give units digit as 8, 9, 0 or 1. It will never be 7.
D. 8 ……. 4*8+(carry over 0-3) will give units digit as 2, 3, 4 or 5. It will never be 8.
E. 9 ……. 4*9+(carry over 0-3) will give units digit as 6, 7, 8 or 9. It can be 9….CORRECT


E
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Easy and Simple.
Thanks! :please:
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If ABCDE*4 = EDCBA. A,B,C,D and E are all natural numbers ( 0-9). what 10 Apr 2025, 14:37
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