Question

How many three digits numbers are there such that sum of the digits is even, number is odd and divisible by 5?

@Directrix @JamesJ

Answer 1

Just checking, my answer is 45.

Answer 2

Hey Ishaan, try it took me 2 mints but my friends think I am wrong :(

Answer 3

Actually she says that there were no option like 45 in the actual test, I have a feeling that she didn't remember the question properly.

Answer 4

sum of the digits is even, numbers are odd and the sum is divisible by 5
or
sum of the digits is even, numbers are odd and numbers are divisible by 5

Answer 5

according to her version,

"sum of the digits is even, numbers are odd and numbers are divisible by 5"

Answer 6

Could you give me one example? I am getting confused :-/

Answer 7

btw, for "sum of the digits is even, numbers are odd and the sum is divisible by 5" is still 45

Answer 8

Okay,


215
235
255
275
295
305
325
345
365
385
415
435
455
475
495
505
525
545
565
585

Answer 9

Okay I wrote a brute-force and seems like I am right! :)

Answer 10

For the case you suggested the numbers are:

109
127
145
163
181
217
235
253
271
299
307
325
343
361
389
415
433
451
479
497
505
523
541
569
587
613
631
659
677
695
703
721
749
767
785
811
839
857
875
893
901
929
947
965
983

Answer 11

Lol Thanks, I thought individual digits must be odd, which made the question impossible. I overthink things :-/

Answer 12

Over-thinking is good, and don't worry I am a Fool myself lol

Answer 13

I am getting 90 in total (Inclusion of numbers ending with ..0 and ..5). So, I think 45 for Odd.

Answer 14

@FoolForMath
Last digit should be 5 to be divisible by 5
If first digit is even, second digit should be odd= \(4\times 5\times 1=20\)
If first digit is odd, second digit should be even=\(5\times 5\times \times 1=25\)
Total three no. digits which satisfy the conditions= 45

Answer 15

How many three digits numbers are there such that sum of the digits is even, number is odd and divisible by 5?

45
assuming no leading zeros in the 3 digit number.
in each decade, there is only 1 number divisible by 5 i.e. ends in 5
in each hundred, there are 10 decades, 5 of which have digits that sum to an even #
so 5 numbers per hundred.
there are 9 hundreds (100, 200,...900), so 9*5= 45

Answer 16

I get 45 as well.

Answer 17

the last digit can be 0 or 5 for the number to be divisible by 5. e.g.
110
130
150
...
so I get 90?

Answer 18

oh - the number must be odd - so it must end in 5 - therefore only 45