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.
Best Answer
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 $(a,b,c)$?
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 $880$.