HOT Competition 3 Sep/8PM: In the correctly worked addition problem sh

Question

ABA
ABC
+ACC
-----------
1416

In the correctly worked addition problem shown, where the sum of the three-digit positive integers ABA, ABC and ACC is 1416, and A, B, and C are different digits, what is the units digit of A*B*C?

A. 9
B. 8
C. 6
D. 4
E. 2

mSKR · Answer

cases:
a+2c= 6
2b+c= 11
3a= 13 - reject

cases:
a+2c= 6
2b+c= 21
3a= 12

so,
a=4
c= 1
b= 10-reject

cases:
a+2c= 16
2b+c=20
3a=12

so,
a= 4
c=6
b= 7

ABC= 4.6.7= 24*7= 168
hence 8 is unit digit - 
hence B

cases:
a+2c= 26
it means a= 9; c= 8
2b+c= 19
hence b = not possible -reject

MrBaby · Answer

I try to find A+A+A=14 first, but no way three times of number can be equal to 14, the correct way is borrow from the ten digit, so I think A should be 4 and borrow 2 more --> (4*3)+2.

Then, B+B+C=21, and A+C+C=6 or 16. Try the number; A+C+C should be 16 when A=4, B=7, C=6.

474
476
+466
-----------
1416

The units digit of A*B*C = 4*7*6 = 168.
  ANS B.

Posted from my mobile device

JonShukhrat · Answer

In the correctly worked addition problem shown, where the sum of the three-digit positive integers ABA, ABC and ACC is 1416, and A, B, and C are different digits, what is the units digit of A*B*C?

We can see that A + A + A =14XX
Hence A cannot be 3 because we will have 9XX
A also cannot be 5 because we will have 15XX

So A must be 4. So we havw 4B4 + 4BC + 4CC
Next we can see that 4XX + 4YY + 4ZZ = 12LL
We need two 1 being carried over to the hundreds digits. Thus the sum of the units and tenth digits must be greater than 10.

So we need 4 + 6 + 6 = X6 and 74 + 76 +66 for X16

So they are 474+476+466

4*6*7 = X8

Hence B

Posted from my mobile device

petrichor · Answer

To add three 3 digit numbers to get 1416, the hundreds digit should be 4 so that the hundreds will contribute 1200 and rest of the numbers will contribute 216.

the unit digits of the three numbers are A + C + C => 4 + C + C

adding three single digit numbers we can get 6, 16 or 26, with our condition of same two numbers and one different number, 6 is not possible (2+2+2).

so it should be one in 
16 => 2+7+7,  4+6+6 , 6+5+5, 8+4+4
26 => 8+9+9

only 4+6+6 matches our 4+C+C criterion.

So the numbers become, 4B4 + 4B6 + 466 = 1416. simply solving yields B=7 .

and the unit digit of A*B*C is 4*7*6 = 16 8

Answer is B

Posted from my mobile device

EgmatQuantExpert · Answer

Given 
 
• ABA
ABC
+ACC
-----------
1416
• A, B, and C are different digits

To Find

• units digit of A×B×C.

Approach and Working Out

• From this we can easily infer that 3A + carry gives us 14.
 o A must be 4.

• BA + BC + CC = 216
 o A = 4
o 2C = 2 or 12.
o C = 1 or 6

• 2B + C = 20 or 21
 o C cannot be 1. (Then B = 10)
o C must be 6.
o 2B = 14
o B = 7

• Numbers are 474, 476 and 466.
 o A×B×C = 4 × 6 × 7 = 168
o Unit digit = 8

Correct Answer: Option B

bumpbot · Answer

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HOT Competition 3 Sep/8PM: In the correctly worked addition problem sh

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