We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

A 100-digit positive integer is divisible by 72. What is the greatest possible value of the sum of the number’s digits?

Help please

Can you give me step by step or an explanation please

IDontKnowMath Feb 17, 2019

#1**+2 **

((10^100) - 64) mod 72 =0

9999999999999999999999999 9999999999999999999999999 9999999999999999999999999 9999999999999999999999936 =100 digits long!

Sum =[98 x 9] + 3 + 6 =891

Guest Feb 17, 2019

#3**+1 **

Mod is short for "Modulus". When you divide a number by the "modulus" it tells you right away if it is divisible by the modulus EVENLY, when the result =0. If it returns any other number, which is the REMAINDER, that means the number does not divide the modulus EVENLY. Examples: 72 mod 7 =2 The remainder. But 77 mod 7=0, which means it divides the modulus evenly with no remainder.

Guest Feb 17, 2019

#5**0 **

Guest, I know what mod is but I do not follow your logic. You lost me right at the beginning.

Where did 64 come from?

I did get the same asnwer as you, so I guess your logic is sound but I cannot follow it.

**Here are my thoughts.**

A 100-digit positive integer is divisible by 72. What is the greatest possible value of the sum of the number’s digits?

72=9*8 So it must be divisable by both 9 and by 8

if a number is divisable by 9 the digits have to add to a multiple of 9

1000 is divisable by 8, any number ending in 000 will be divisable by 8.

so the first 97 digits could be 9

now I have to think about the units, 10s and 100 places.

The sum of the 3 digits must be add to 9,18 or 27. (in order for the number to be disable by 9)

27 is no good because the number would end in 9 and so not be divisable by 8

How about 18

864 that works

so I am going to say that the greatest possible value for the sum of the digits is

**9*97+18 = 891**

Melody Feb 18, 2019

#6**+1 **

Hi Melody: He/she wants a 100-digit number which is divisible by 72.

I simply tried: 1E100 mod 72 =64 as the remainder. So, if I subtracted it from 1E100, the number would consist of all 9s and 36 and the end and be divisible by 72 with no remainder, which is the case. At the same time, it gives him/her the largest sum of the 100 digits, 98 of them being 9s and the last two being 36.

Guest Feb 18, 2019

#7**0 **

ok thanks I get it.

You had to use a calc for that of course. :)

10^100 mod 72 = 64 I just used the web2.0 calc for that.

So the biggest number possible will be 10^100 - 64

the smallest 2 digits will be 36 and all the rest will be 9.

9*98+9 = 891.

Ok I can see what you have done and I can see it has worked this time but

I question if it would work everytime.

--------

I am just thinking out loud now.

You have found the biggest number, but I am nost so sure that the biggest number would always have the greatest sum of the digits for a question like this.

I mean I can see that it __has__ to work for this particular example because

the 2 smallest have to add up to 9, or

the 3 smallest have to add to 18 or

the 4 smallest have to add to 27 etc

And of course those small numbers can be placed anyway, they do not have to have the smallest place value.

but for other examples.......not so sure.

eg

997>989 however 9+9+7<9+8+9

**I do like your logic though - thanks for sharing it :)**

Melody
Feb 18, 2019