Question

Find the smallest number that is divisible by both 3 and 7 and also the sum of the digits of this number is divisible by 3 and 7. For example, 42 is divisible by 3 and 7 but 4+2=6 is not divisible by 7. Therefore, 42 is not the answer. The answer is a 3-digit number less than 500.

Answer 1

well, it has to be a multiple of 21 for sure

Answer 2

21^2 by chance?

Answer 3

nah, 441 is 9

Answer 4

21x = 100a+10b+c
a+b+c = 21y

just another random thought

Answer 5

arent all 3 digit numbers less than 5000?

Answer 6

sorry, its 500, not 5000, dumb computer XD thank you for helping :) this is for my shsat help so i can get into a good highschool

Answer 7

none of the numbers are going to add to anyting much greater than 21

499 =22 at best

Answer 8

so we have the number is a 3 digit number
so we know we have p=21*k for some integer k>0 (i'm assuming smallest positive number of course)

we also have p=d_2d_1d_0

so we have
\[d_2 10^2+d_110^1+d_010^0=21 \cdot k \]

we also have
\[d_210^2+d_110^1+d_010^0=3 k_1 \]
so we know
\[d_1+d_2+d_3=3 \cdot k_2\]

I don't know if there is a trick for 7.

Also I put different k's above because they aren't equal k's

Answer 9

actually you don't have to go far in the 100's to find the answer

Answer 10

actually the fist number that takes in the 3 digits numbers is 21*5
and guess what?

Answer 11

oops sum of digits sorry

Answer 12

105 is divisible for 3 and 7

i forgot it said digits of the number

Answer 13

21x = 100a+10b+c
21 = a+b+c

21 = (100a+10b+c)/x

a+b+c = (100a+10b+c)/x

ax+bx+cx = 100a+10b+c

0 = (100-x)a+(10-x)b+(1-x)c

maybe we can go with another thought .............

21x = 100a+10b+c
21 = a+b+c

x = 100a + 10b + c
--------------
a+b+c

the only way i can think of is to make sure is to brute math it lol

Answer 14

there are only 21 numbers to deal with that are less than 500

Answer 15

21=3
42=6
63=9
84=12
105=6
126=9
147=12
168=15
189=18
210=3
231=6
252=9
273=12
294=15
315=9
336=12
357=15
378=18
399=21 ..... this is it
420
441
462
483

Answer 16

fine 23 numbers lol

Answer 17

yeah subtracting both equations eliminates c and the problem simplifies to :

100a+10b+c = 21k
a+b+c = 21y
----------------------
99a+9b = 21m

Answer 18

y = 1