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If abc is a three digit number such that not one number is similar to other than

1 How many possible values of ( a + 4b + c ) will be divisible by 40?

2 How many possible values will be divisible by 40 :

My approach: i could solve only second question

For 40, last 3 digit number must be divisible by 8 and 5 For this to be possible

Possible nos are:

1 120

2 160

3 240

4 280

5 320

6 360

7 480

8 520

9 560

10 640

11 680

12 720

13 760

14 840

15 920

16 960

So,16 ways are there for which the number is divisible by 40.

Is there any better approach for solving the second problem? plus a hint for solving first problem.

Jack
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2 Answers2

2

  1. You have $0 \leq b,c \leq 9$ and $1\leq a \leq 9$, then $1 \leq a+4b+c \leq 9*6 = 54$. So, if $a+4b+c$ is divisible by $40$, then $a+4b+c = 40$. Now, can you find such $(a,b,c)$?

  2. Note that from $100$ to $1000$, there are $\left[ \frac{900}{40}\right] = 22$ numbers which is divisible by 40. But we have $200$, $400$, $440$, $600$, $800$ and $880$.

GAVD
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Outline Part A:

In order for $(a + 4b + c)$ to be divisible by $40$, $a + 4b + c = 40$ (cannot be 80 or more), $b$ must be at least $6$. For $b = 6$, we can have $(a, c)$ = $(7,9), (9,7)$ as two possibilities. For $b = 7$, you have $(a, c)$ = $(4,8), (8,4), (3,9), (9,3)$ as $4$ possibilities. Can you work out the possibilities for $b = 8$ and $b = 9$ ? . The final answer should be $2 + 4 + 6 + 3 = 15$ possibilities.

Outline Part B:

In this case, can you see that $c = 0$, and the problem reduces to all $ab$'s which are divisible by $4$ ?

Shailesh
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