A few days ago
SapphireB

Weekend Family Math?

I am trying to help my step-son with his math.

Problem of the weekend:

(Be sure to explain how you got your answer.)

****Find the LARGEST six-digit even number that meets the following conditions. The ten-thousands digit is twice the tens digit. The hundred thousands digit is a prime number. The thousands digit is divisible by the ones digit. NO digits can used more than once.

Thank you for helping us out.

Top 4 Answers
A few days ago
Anonymous

Favorite Answer

I got 789, 643.

I started out as _ _ _,_ _ _

The 10,000 place has to be the biggest you can get. However, 9 isn’t an even number so I used 8., so the 10’s digit is a 4. The biggest prime number from 1-10 is 7, for the hundred thousands place. The thousands digit has to be the biggest number divisible by another number, which is 9, so the 1’s digit is a 3. The only place left is the hundreds place, and I used the biggest number i can get, 6.

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A few days ago
TurtleFromQuebec
Call the number N = ABCDEF, where A is the hundred thousands digit and down to F as the units digit. The trick is to think “largest” at each step, this tells us to apply the conditions left to right.

A must be prime, so A can be 2,3,5,7. Choose biggest: A = 7.

B is twice E, so it must be even, Let B = 8, biggest even digit, so E is 4. Now we have N = 78CD4F.

C is divisible by F, make C as big as possible for C = 9. Now N = 789D4F.

D can be anything, biggest remaining digit is 6.

F must be a divisor of C, biggest one is 3.

So it looks like N = 789643.

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A few days ago
cloud43
Try 786,941

8 is twice 4

7 is a prime number

6 is divisible by 1 ( 1 goes into 6 six times.)

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4 years ago
Anonymous
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