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# Hmmmm

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I cant solve this. Lol CU can u?

There are two 3-digit numbers N having the property that N is divisible by 11 and \(\dfrac{N}{11} \) is equal to the sum of the squares of the digits of N.

Find both values of N You may submit them in either order.

Mar 23, 2020

#1
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Mar 23, 2020
edited by ElectricPavlov  Mar 23, 2020
edited by ElectricPavlov  Mar 23, 2020
edited by ElectricPavlov  Mar 23, 2020
#2
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Sorry I'm not CU, but I'll try this question. I've done a variant of it before I think.

Let's call N = 100a + 10b + c

Then we know:

100a + 10b + c = 11m for some arbitrary integer m

m = a^2 + b^2 + c^2

100a + 10b + c = 11a^2 + 11b^2 + 11c^2.

Using divisibility rule of 11, we know:

Case 1:

a-b+c = 0

a+c = b

or Case 2:

a-b+c = 11

b = a+c-11

this splits the problem into two cases for us:

1. b = a +c where the first case for divisibility of 11 holds true.

100a + c + 10a + 10c = 11a^2 + 11c^2 + 11(a+c)^2

10a + c = 2a^2 + 2ac+ 2c^2

10a + c = 2a^2 + 2ac + 2c^2

Here, we're kind of stuck. Realize that we can factor out 2 from the right hand side, meaning that c must be even since a is also even.

we can write c = 2z, with z being either 0, 1, 2, 3, or 4, and substitute it in

we get:

10 a + 2z = 2a^2 + 4az + 8q^2

Dividing by 2 on both sides, this gives us:

5a + z = a^2 + 2az + 4z^2

We just bash out cases now, and eventually we find:

when we have z = 0, the case works out, with a being 5, and c = 0  b = 5. From this case, we get that N = 550. All other cases don't work(2,4,6,8)

I'll leave the rest of the problem to you; there's only one other case!

2. ?????

also edit: sorry for the bad "typography". Had to do this in a hurry

Mar 23, 2020
edited by jfan17  Mar 23, 2020
edited by jfan17  Mar 23, 2020
#3
+23566
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Good work, jf !

ElectricPavlov  Mar 23, 2020
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Kudos to you to!

jfan17  Mar 23, 2020
#5
+483
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Also just wanted to put this out there. If you have lots of time to do this problem, I'd actually recommend just bashing out all of the possibilities divisible by 11. Because there are only ~ 80 - 81(do the math?) divisors of 11 from 100 - 999, I would recommend listing all of them out and then testing on a case by case basis(of course, this heavily relies on you having lots of time).

Mar 23, 2020
#6
+1956
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Let N=110a+10b+c for some digits a,b, and c . Thenfor some .

We also have m=a^2+b^2+c^2.

For an integer divisible by 11, the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by 11.

Substituting, we get 100a+10b+c=11a^2+11b^2+11c^2

Case 1: b=1+c.

100a+c+10a+10c=11a^2+11c^2+11(a+c)^2.

10a+c=2a^2+2ac+2c^2

Case 2: Substitute q= 0-4 and solve for a

q=0: This works.

q=1: This is not an integer. This doesn't work for a.

q=2: This is not an integer. This doesn't work for a.

q=3: This is not an integer. This doesn't work for a.

q=4: This is not an integer. This doesn't work for a.

Another solution: There are 900 three-digit numbers. Only 81 of these numbers are divisible by 11. Now, list the answer. We end up getting N=550,803.

Hope this helped!

Mar 23, 2020
#7
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Impressive work guys    !!!!

CPhill  Mar 23, 2020
#8
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Welp I'm late I just woke up 3 hours ago and its 2:30 at my location. Nice work guys

Mar 23, 2020
#9
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Thank you!!

Mar 23, 2020