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If A, B and C represent different digits in the multiplicati

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If A, B and C represent different digits in the multiplicati [#permalink] New post 30 Jun 2018, 10:00
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If A, B and C represent different digits in the multiplication, then A + B + C =
Attachment:
multiplication.png
multiplication.png [ 905 Bytes | Viewed 4953 times ]


A 9
B 12
C 14
D 15
E 17
[Reveal] Spoiler: OA
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Re: If A, B and C represent different digits in the multiplicati [#permalink] New post 01 Jul 2018, 12:39
it is E...had to spend a lot of time figuring it out...took 335, 551, 553, 445, 776...atlast 776 worked :shock: :roll:
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Re: If A, B and C represent different digits in the multiplicati [#permalink] New post 01 Jul 2018, 12:40
Please do share the shortcut if you have as solution...thanks
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Re: If A, B and C represent different digits in the multiplicati [#permalink] New post 02 Jul 2018, 07:08
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anuppatel012 wrote:
Please do share the shortcut if you have as solution...thanks


It is a step by step process,


First we have to calculate for Unit's digit B

We need a number that is B*B = B

So let us start as follows:

STEP 1

a) 0∗0=0 But entire answer will become zero

b) 1∗1=1 but the product will remain as AAB

c) 5∗5=25 but that won't be possible as (5 * A) + 2 = 7 , but we need 5

d) 6∗6=36 and 6∗A+3=5, so this value is Possible when A is 2 or A is 7

no other possible solution, so B is 6 and A is 2 or 7

STEP II

The product is AA6∗6=C656

Now if A is 2, it becomes 226∗6=1356 ; but we need C656

so if A is 7

776∗6=4656 which is = CB5B

Hence A+B+C=7+6+4=17
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Re: If A, B and C represent different digits in the multiplicati [#permalink] New post 14 Nov 2019, 07:47
Official Explanation from Magoosh

AAB × B = CB5B

The first thing we should note is the role of B—it's the units digit of both factors and of the product. So, it must be a number that we can square and result in a units digit that equals the number that we squared. That's a mouthful. Let's see some examples.

0 * 0 = 0
1 * 1 = 1

This is a very common trick on variables-for-digits questions like this, and it's good to know (without doing the math) that the four possible digits that result in a number with the same units digit when squared are 0, 1, 5, and 6.

If you didn't know that immediately, then check:

2 * 2 = 4 [no good]

3 * 3 = 9 [no good]

4 * 4 = 16 [no good]

5 * 5 = 25

6 * 6 = 36

7 * 7 = 49 [no good]

8 * 8 = 64 [no good]

9 * 9 = 81 [no good]

At this point we have four numbers to test. B = 0 is impossible—CB5B would have to equal 0. Similarly, B = 1 is impossible, because then the product would have to be a three digit number equal to AAB, not a four digit number.

So what if B = 5? When we multiply those 5s in the units, we get 25. So that works. Now carry the 2 next to the A. Now we know (5 * A) + 2 that again results in 5. No value of A is possible for this; well will always end up with a 7 or a 2. For example,

(5 * 2) + 2 = 12

(5 * 3) + 2 = 17

So B can't be 5. By the process of elimination, B must equal 6!

AA6 × 6 = C656

Now, given B = 6, so 6 * 6 = 36. That means (6 * A) + 3 must give us a number that ends in 5. Let's experiment,

(6 * 2) + 3 = 15

(6 * 3) + 3 = 21 [no good]

(6 * 4) + 3 = 27 [no good]

(6 * 7) + 3 = 45

That means that either A = 2 or A = 7. When we add that 3 (carried from 6 * 6 = 36), we get a tens digit in the product that equals 5.

It turns out that A = 2 doesn't work because of the A in the hundreds place and the B in the product. If A were 2 then B would have to be 3. For example,

(2 * 6) + 1 = 13

We already know that B must be 6 so we can eliminate 2 as a possibility. So A = 7 and we can solve the problem.

A = 7, B = 6, and C = 4, so A + B + C = 17

Answer = (E)

In case you're curious, with all the numbers in place, the product is

766 × 6 = 4656

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Re: If A, B and C represent different digits in the multiplicati [#permalink] New post 18 Nov 2019, 21:56
amorphous wrote:
If A, B and C represent different digits in the multiplication, then A + B + C =
Attachment:
multiplication.png


A 9
B 12
C 14
D 15
E 17


There are only 2 possible numbers which result in the same unit digit when multiplied by itself.
Those numbers are 5 and 6.
Now 5*5 = 25 and 6*6 = 36.

if the unit digit is 5 then the carry over is 2. and BA + 2 = 5. and BA = 3.
now in the multiplication table of 5 the unit digits possible are 0 and 5. and when 2 is added to it the unit digit can be either 2 or 7.
so 5 is ruled out.

now, B = 6. and 6A + 3 = 5 so 6A = 2 as the unit digit.
now the only possible value of A is either 2 or 7.
as 6*2 = 12(unit digit 2) and 6*7 = 42(unit digit 2).
but if we take A as 2 then in the next multiplication i.e. 6*A = 6 as the unit digit will not be possible.
as 226 * 6 results in 1356.

So, A = 7.
and the final multiplication becomes 776 * 6 = 4656
and A + B + C = 17

OA - E
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Re: If A, B and C represent different digits in the multiplicati [#permalink] New post 19 Nov 2019, 10:35
THis is a tough one. We have to proceed by trial and error.
For B, only two integers are possible, 5 and 6 because of 5*5 = 25, the unit is equal to 5
and 6*6 = 36 (unit of 36 equal to 6.
Then, you have to rule out 5 because 5 times another integer gives an integer with the unit equal to 0 or 5.
So we are left with 6. To get 5 of the number CB5B, we have 7 (6*7=42, add 3 to 42 and you get 45).
A = 7. B = 6 and C = 4
A+B+C=7+6+5 = 17
SOLUTION: E
Re: If A, B and C represent different digits in the multiplicati   [#permalink] 19 Nov 2019, 10:35
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