Q: Let abc be a number with 3 digits. Similary let cba be a number with 3 digits. How many numbers exist such that abc is divisible by 3 and cba is divisible by 5
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graceful hare
solar plinthBOT
graceful hare
My work: If cba is divisible by 5, then a has to be 5 or 0. But since a cannot be 0 bc it's a number with 3 numerals, a is 5. Now plug it in and we get the following condition: $5+10b+c \equiv 0$ mod 3. Bc $10 \equiv 1$ mod 3 we can replace 10 with 1 and we get $5+b+c \equiv 0 $ mod 3
silk veldtBOT
Tony
graceful hare
And I also calulated how many numbers with 3 digits have are divisible by 3 and 5
First we need to find the least common multiple of 3 and 5 which is 15
Now all 3 digit numbers are between 100 and 999
so we need to calculate $\floor*{\frac{100}{15}}$
silk veldtBOT
Tony
graceful hare
this is 6
so there are 6 numbers with 2 digits having the divisble prperty
and then calculate: $\floor*{\frac{999}{15}}$
silk veldtBOT
Tony
graceful hare
this is 66
now subtract them and we get 60
there are 60 3 digit numbers that are divisible by 5 and 3
idk if thats useful
zealous notch
Q: Let abc be a number with 3 digits. Similary let cba be a number with 3 digits...
if abc is divisible by 3, bca bac acb cba cab are also divisible by 3. The order doesn’t matter.
If a = 5
and (a + b + c) is divisible by three
That means b+c can be 4, 7, 10, 13, 16
We know that a ≠ 0 and c ≠ 0
So the conditions are all enumerated above
b + c = 4
When b = 0~3, there is a counterpart c
b + c = 7
b = 0~6
b + c = 10
b = 0~9
etc etc
Chinese remainder isn’t necessary here
graceful hare
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