# 8 digit number

• Feb 21st 2013, 02:55 AM
RuyHayabusa
8 digit number
Find a 8 digit number which starts with 1984 and is divisible by 6,7,8,9.Step by step explanation will be appreciated.
• Feb 21st 2013, 06:12 AM
HallsofIvy
Re: 8 digit number
So the number is . Any number divisible by both 8 and 9 is necessarily divisble by 6 so we really only need to check "divisible by 8", divisible by 7" and "divisible by 9".

To be divisible by 8, the first three digits must make a number divisible by : 008, 016, 024, 032, 040, 048, etc.

To be divisible by 9, the sum of the digits must be divisible by 9: 1+9+ 8+ 4= 22 so 22+ a+ b+ c+ d must be divisible by 9.
22+ 0+ 0+ 8= 30
22+ 0+ 1+ 6= 29
22+ 0+ 2+ 4= 28
22+ 0+ 3+ 2= 27 is divisible by 9 so 1984a032 is a candidat
22+ 0+ 4+ 0= 26
22+ 0+ 4+ 8= 34
22+ 0+ 5+ 6= 33
22+ 0+ 6+ 4= 32
22+ 0+ 7+ 2= 31
22+ 0+ 8+ 0= 30
22+ 0+ 8+ 8= 38
22+ 0+ 9+ 6= 37
22+ 1+ 0+ 4= 27 so 1984a104 is a candidate, etc.

A number is divisible by 7 if and only if after taking the last digit off the number, doubling that last digit and subtracting from the truncated number, the result is divisible by 7. For example 805 is divisible by 7 because we drop the last digit, 5, we have 80. Subtracting 2(5)= 10 from that gives 70 which is obviously divisible by 7.
• Feb 21st 2013, 06:58 AM
zendo
Re: 8 digit number
I multiplied 6*7*8*9 together to get 3024.

I know that i can multiply any number by 3024 and it will be divisible by 6, 7, 8, or 9.

then by trial and error (2 mintues) i zeroed in on a number to multiply 3024 by to get the required number. 3024 * 6563 to get 19846512.
• Feb 23rd 2013, 02:43 AM
RuyHayabusa
Re: 8 digit number
Quote:

Originally Posted by HallsofIvy
So the number is . Any number divisible by both 8 and 9 is necessarily divisble by 6 so we really only need to check "divisible by 8", divisible by 7" and "divisible by 9".

To be divisible by 8, the first three digits must make a number divisible by : 008, 016, 024, 032, 040, 048, etc.

To be divisible by 9, the sum of the digits must be divisible by 9: 1+9+ 8+ 4= 22 so 22+ a+ b+ c+ d must be divisible by 9.
22+ 0+ 0+ 8= 30
22+ 0+ 1+ 6= 29
22+ 0+ 2+ 4= 28
22+ 0+ 3+ 2= 27 is divisible by 9 so 1984a032 is a candidat
22+ 0+ 4+ 0= 26
22+ 0+ 4+ 8= 34
22+ 0+ 5+ 6= 33
22+ 0+ 6+ 4= 32
22+ 0+ 7+ 2= 31
22+ 0+ 8+ 0= 30
22+ 0+ 8+ 8= 38
22+ 0+ 9+ 6= 37
22+ 1+ 0+ 4= 27 so 1984a104 is a candidate, etc.

A number is divisible by 7 if and only if after taking the last digit off the number, doubling that last digit and subtracting from the truncated number, the result is divisible by 7. For example 805 is divisible by 7 because we drop the last digit, 5, we have 80. Subtracting 2(5)= 10 from that gives 70 which is obviously divisible by 7.

Thanks!