Saturday

October 25, 2014

October 25, 2014

Posted by **Ryan** on Wednesday, September 30, 2009 at 9:59pm.

- Pre-Calc (ugent!!!) -
**seth**, Wednesday, September 30, 2009 at 10:07pmhaha is this a last minute calendar problem by any chance? because it's october tomorrow and i have have this problem too! lets hope somebody answers soon...

- Pre-Calc (ugent!!!) -
**MathMate**, Wednesday, September 30, 2009 at 10:11pmThe digit combinations of a 5-digit number having a sum of 43 is quite limited.

The digits are 99997, 99988.

99997 can place the 7 in 5 positions, so the number of outcomes using 4-9's and a 7 is 5.

Using 3-9's and 2-8's has 5!/(3!2!)=10 ways of arranging the numbers.

Of these, only 98989, 97999 and 99979 are divisible by 11.

Can you continue?

Check my thinking.

- Pre-Calc (ugent!!!) -
**Ryan**, Thursday, October 1, 2009 at 6:36pmYes seth it is

**Answer this Question**

**Related Questions**

Maths - What are 3 5-digit numbers that have the sum of their digits equal? Is ...

Pre-Algebra - -My tens and units digits are consecutive integers whose product ...

maths - Find the sum of all 3-digit positive numbers N that satisfy the ...

math - You have 2 digits, 2 numbers, reverse digits and 54 if the difference and...

Finite Math - Assume that 3 digits are selected at random from the set {1,3,5,6,...

Pre Calc - the units digit of a 2 digit number is 4 less than the tens digit. ...

Algebra - Let S(N) denote the digit sum of the integer N. Let M denote the ...

math - I am a 4-didit number. All my digits are odd numbers and each digit is ...

Maths - The Sentosa High School's telephone number is an eight digit number. The...

math - Assume that 3 digits are selected at random from the set { 2, 5, 6, 8, 9...