# Maths Problem

Watch
Announcements

This discussion is closed.

Report

#21

(Original post by

I get the gist of this but the step in bold, where's it from and what's it mean?

**ZJuwelH**)I get the gist of this but the step in bold, where's it from and what's it mean?

10a + b = 11a + 11b

11a + 11b - 10a -b = 0

11a - 10a + 11b - b = 0

a + 10b = 0

(it means either a and b are both 0, or a and b are of opposite sign, which cannot be.).

0

Report

#22

(Original post by

10a + b = 11(a + b)

10a + b = 11a + 11b

11a + 11b - 10a -b = 0

11a - 10a + 11b - b = 0

a + 10b = 0

(it means either a and b are both 0, or a and b are of opposite sign, which cannot be.).

**elpaw**)10a + b = 11(a + b)

10a + b = 11a + 11b

11a + 11b - 10a -b = 0

11a - 10a + 11b - b = 0

a + 10b = 0

(it means either a and b are both 0, or a and b are of opposite sign, which cannot be.).

But the one-digit number 0 would work

0

Report

#23

(Original post by

Duuuh <slaps forehead> and I want to study Maths at Cambridge...

But the one-digit number 0 would work

**ZJuwelH**)Duuuh <slaps forehead> and I want to study Maths at Cambridge...

But the one-digit number 0 would work

0

Report

#24

(Original post by

But the one-digit number 0 would work

**ZJuwelH**)But the one-digit number 0 would work

0

Report

#25

(Original post by

You have to start by proving you can only do this when the number in question is a 3 digit number. Then let n (the 3 digit number) = 100a + 10b + c where a,b,c are integers <10 and then set that equal to 11(a+b+c) (i.e. 11 times the sum of their digits)

Let me know if you're stuck again.

**theone**)You have to start by proving you can only do this when the number in question is a 3 digit number. Then let n (the 3 digit number) = 100a + 10b + c where a,b,c are integers <10 and then set that equal to 11(a+b+c) (i.e. 11 times the sum of their digits)

Let me know if you're stuck again.

100a+10b+c = 11a+11b+11c

89a = b+10c (the most constructive rearrangement of this equation I could find, and even then I'm stuck)...

Sorry theone I flopped your challenge, looks like it's Manchester for me!

0

Report

#26

(Original post by

n = 100a+10b+c = 11(a+b+c)

100a+10b+c = 11a+11b+11c

89a = b+10c (the most constructive rearrangement of this equation I could find, and even then I'm stuck)...

Sorry theone I flopped your challenge, looks like it's Manchester for me!

**ZJuwelH**)n = 100a+10b+c = 11(a+b+c)

100a+10b+c = 11a+11b+11c

89a = b+10c (the most constructive rearrangement of this equation I could find, and even then I'm stuck)...

Sorry theone I flopped your challenge, looks like it's Manchester for me!

0

Report

#27

(Original post by

there is a special condition that links a and c, but i've forgotten it. i think this question was in the STEP 1 paper i took in the summer.

**elpaw**)there is a special condition that links a and c, but i've forgotten it. i think this question was in the STEP 1 paper i took in the summer.

What about numbers greater than 3 digits...

0

Report

#28

(Original post by

Well it's clear a can only be 1 and then b and c must possess a certain value?

What about numbers greater than 3 digits...

**theone**)Well it's clear a can only be 1 and then b and c must possess a certain value?

What about numbers greater than 3 digits...

0

Report

#29

(Original post by

Sorry it's not clear to me why a = 1, is it because otherwise the number isn't three digits? Not at my mental best, bloody Chelsea how dare they

**ZJuwelH**)Sorry it's not clear to me why a = 1, is it because otherwise the number isn't three digits? Not at my mental best, bloody Chelsea how dare they

Now b and c <10 so the RHS < 100 so 89a < 100 so a must be 1.

0

Report

#30

(Original post by

We have 89a = 10b + c agreed?

Now b and c <10 so the RHS < 100 so 89a < 100 so a must be 1.

**theone**)We have 89a = 10b + c agreed?

Now b and c <10 so the RHS < 100 so 89a < 100 so a must be 1.

0

(Original post by

Duuuh again. At this rate I won't even get in to Manchester, it's Queen Mary for me! Still, I'm going to try this,...

**ZJuwelH**)Duuuh again. At this rate I won't even get in to Manchester, it's Queen Mary for me! Still, I'm going to try this,...

0

Report

#32

So if 89a = 10b+c and a = 1:

89 = 10b + c

But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

So far I have 0 and 189 yay!

89 = 10b + c

But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

So far I have 0 and 189 yay!

0

Report

#33

(Original post by

since that is the case, then c must be 8. And b must be 9.

**2776**)since that is the case, then c must be 8. And b must be 9.

but there could be other (4+ digit) answers....

0

(Original post by

So if 89a = 10b+c and a = 1:

89 = 10b + c

But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

So far I have 0 and 189 yay!

**ZJuwelH**)So if 89a = 10b+c and a = 1:

89 = 10b + c

But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

So far I have 0 and 189 yay!

0

Report

#35

**ZJuwelH**)

So if 89a = 10b+c and a = 1:

89 = 10b + c

But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

So far I have 0 and 189 yay!

0

Report

#36

**ZJuwelH**)

So if 89a = 10b+c and a = 1:

89 = 10b + c

But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

So far I have 0 and 189 yay!

0

Report

#37

(Original post by

so 198 = 11 (1 + 9 + 8) = 198 !

but there could be other (4+ digit) answers....

**elpaw**)so 198 = 11 (1 + 9 + 8) = 198 !

but there could be other (4+ digit) answers....

0

Report

#40

**elpaw**)

so 198 = 11 (1 + 9 + 8) = 198 !

but there could be other (4+ digit) answers....

4 digit numbers gives:

989a + 89b = 10d + c

Which is impossible for a,b,c,d < 10 because max of 10d + c is 99 but the minimum of the lhs is 989 because a has to be at least 1

0

X

new posts

Back

to top

to top