Look at Kalva... Bomb

This how I did it. (Sorry, it's not IMO-style and is really long.) Every number that is divisible by 11 has this property: the sum of digits in the odd positions minus the sum of digits in the even positions is a muliptle of eleven, but since N is a 3-digit number there are only 2 cases: b = a + c or b = a + c - 11. Therefore: Now that we have 2 quadratic equations, we substitute for all the possible a's (since a cannot be 0). But since it's a quadratic equation, the discriminant has to be positive. Therefore, we could check whether the discriminant is positive to save us time. Plugging in each possible a , we see that first equation only works when a is equal to 1, 2, 3, 4, 5 and 6, and the second works when a is 5, 6, 7 and 8. Then we solve all the quadratic equations for integral values, which gives us 550 and 803. The second equation, when a was equal to 5, returned 3 as one of its roots, which would have given us 583, but since b = a + c - 11 and 5 + 3 -11 = -3, it does not work. Thanks, Rust, for catching my mistake. edit: I just realized I used a, b and c for different things, which makes it kind of confusing. Sorry!

emilgouliev wrote: Every number that is divisible by 11 has this property: the sum of digits in the odd positions is equal to the sum of digits in the even positions. Therefore: b=a+c It is not true. For example 11|616 (b=a+c-11). However this equation had not another solutions.

Oh, right! I didn't really check it and remembered it wrong. I'll try to fix my solution. Thanks!

I don´t know if bringing back to life ancient topics is against some rule here... Is it? Well, independently of that, I think it´s a good idea to have all IMO problems solved here. So I´ll give some contribuitions (while I can do the problems!). In this problem, particularly, I don´t think it´s a good idea to use the brain. Isn´t it simple to just test all cases? If it is... Let´s go. Let m = N/11 and S the sum of the digits of N. I am not going to make the calculations when S becomes greater tham m, because S grows more, except if 9 comes 0, for example from 319 to 330. However, for 660 or more, we have sum 36 + 36 = 72, that is sufficiently big (for greater m, we have greater S). N = 110 gives m = 10 and S = 2. N = 121 gives m = 11 and S = 6. N = 132 gives m = 12 and S = 14. N = 143 gives m = 13 and S = 26. N = 154 gives m = 14 and S = 42. . . . N = 209 gives m = 19 and S = 85. N = 220 gives m = 20 and S = 8. N = 231 gives m = 21 and S = 14. N = 242 gives m = 22 and S = 24. . . . N = 308 gives m = 28 and S = 73. N = 319 gives m = 29 and S = 91. N = 330 gives m = 30 and S = 18. N = 341 gives m = 31 and S = 26 N = 352 gives m = 32 and S = 38. . . . N = 407 gives m = 37 and S = 65. N = 418 gives m = 38 and S = 81. N = 429 gives m = 39 and S = 101. N = 440 gives m = 40 and S = 32. N = 451 gives m = 41 and S = 42. . . . N = 506 gives m = 46 and S = 61. N = 517 gives m = 47 and S = 75. N = 528 gives m = 48 and S = 93. N = 539 gives m = 49 and S = 115. N = 550 gives m = 50 and S = 50. . . . N = 605 gives m = 55 and S = 61. . . . N = 704 gives m = 64 and S = 65. N = 715 gives m = 65 and S = 75. . . . N = 803 gives m = 73 and S = 73. . . . N = 902 gives m = 82 and S = 85. . . . And end. Obs.: when I asked if I am allowed to revive this topic, it was not a retoric question; please, answer.

Sorry for the revision of old problems.

DPopov wrote: Determine all three-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 $N$. See this

mathmdmb wrote: DPopov wrote: Determine all three-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 $N$. See this lols i added my solution the that page

We let $N=100a+10b+c=99a+11b+a-b+c=11a^2+11b^2+11c^2$. We know that $99a+11b$ is divisible by $11$ so $a-b+c\equiv 0\pmod{11}$. We also know that $0\le a+c\le 18$ because $a$ and $c$ are digits so $b=a+c$ or $b=a+c-11$ since $b$ is also a digit. We know that if $b=a+c$ then \begin{align*} 100a+10b+c &= 11(a^2+b^2+c^2) \\ 110a+11c &= 11(a^2+(a+c)^2+c^2) \\ 10a+c &= 2a^2+2ac+2c^2 \\ 0 &= 2a^2+(2c-10)a+(2c^2-c). \\ \end{align*} We take $0 = 2a^2+(2c-10)a+(2c^2-c)$ modulo $2$ to get $0\equiv c\pmod2$. We also know that the discriminant has to be non-negative so \begin{align*} (2c-10)^2-4\cdot2(2c^2-c)&\ge0 \\ (c-5)^2-2(2c^2-c)&\ge0 \\ c^2-10c+25-4c^2+2c&\ge0 \\ -3c^2-8c+25\ge0 \\ \end{align*}and $-5<c<2$ (not strict) so the only even non-negative integer that is possible is $c=0$. We substitute $c=0$ into $0 = 2a^2+(2c-10)a+(2c^2-c)$ to get $a=5$. We know that $b=a+c$ so $b=5$ and $N=550$. We also have to consider the case $b=a+c-11$. We know that if $b=a+c-11$ then \begin{align*} 100a+10b+c &= 11(a^2+b^2+c^2) \\ 110a+11c-110 &= 11(a^2+(a+c-11)^2+c^2) \\ 10a+c-10 &= 2a^2+2ac-22a-22c+2c^2+121 \\ 0 &= 2a^2+(2c-32)a+(2c^2-23c+131). \\ \end{align*} We take $0 = 2a^2+(2c-32)a+(2c^2-23c+131)$ modulo $2$ to get $1\equiv c\pmod2$. We also know that the discriminant has to be non-negative so \begin{align*} (2c-32)^2-4\cdot2(2c^2-23c+131)&\ge0 \\ (c-16)^2-2(2c^2-c+10)&\ge0 \\ c^2-32c+256-4c^2+2c-20&\ge0 \\ -3c^2-30c+236\ge0 \\ \end{align*}and $-16<c<6$ (not strict) so the only non-negative odd integers that work are $1$, $3$, and $5$. We can test all three of these \begin{tabular}{c|c|c} $c$ & $a$ & $b$ \\ \hline $1$ & not int & not int \\ \hline $3$ & $8, \cancel{5}$ & $0, \cancel{-3}$ \\ \hline $5$ & not int & not int \\ \end{tabular}to get that $N=803$ is the only possible value when $b=a+c-11$. Thus, $N=550$ or $N=803$.

for n in range(110,990,11): unitsDigit=n%10 tensDigit=(n//10)%10 hundredsDigit=n//100 if n//11==hundredsDigit**2+tensDigit**2+unitsDigit**2: print(n)for n in range(110,990,11): unitsDigit=n%10 tensDigit=(n//10)%10 hundredsDigit=n//100 if n//11==hundredsDigit**2+tensDigit**2+unitsDigit**2: print(n)RunResetPop Out /

Replace $N$ with $n$. Assume $n$ is always a three-digit multiple of $11$. Case 1: $n$ has leading digit $1$. $n=110=11\cdot 10$ doesn't work as $1^2+1^2+0^2=2$. $n=121=11\cdot 11$ doesn't work as $1^2+2^2+1^2=6$. $n=132=11\cdot 12$ doesn't work as $1^2+3^2+2^2=14$. $n=143=11\cdot 13$ doesn't work as $1^2+4^2+3^2>13$. The values of $n$ with leading digit $1$ that are multiples of $11$ and greater than $143$ all have a digit greater than or equal to $5$. But $\frac{n}{11}<25$, a contradiction. So nothing with leading digit $1$ works. Case 2: $n$ has leading digit $2$. $n=209=11\cdot 19$ doesn't work as $9^2>19$ $n=220=11\cdot 20$ doesn't work as $2^2+2^2+0^2=8$. $n=231=11\cdot 21$ doesn't work as $2^2+3^2+1^2=14$. $n=242=11\cdot 22$ doesn't work as all the squares we are adding are multiples of $4$. $n=253=11\cdot 23$ doesn't work as $5^2>23$. If $n$ has leading digit $2$ and greater than $253$, it must have a digit greater than or equal to $6$, but $\frac{n}{11}<36$. So nothing with leading digit $2$ works. Case 3: $n$ has leading digit $3$. $308=11\cdot 28$ doesn't work as $8^2>28$. $319=11\cdot 29$ doesn't work as $9^2>29$. $330=11\cdot 30$ doesn't work as $3^2+3^2+0^2=18$. $341=11\cdot 31$ doesn't work as $3^2+4^2+1^2=26$. $352=11\cdot 32$ doesn't work as $3^2+5^2>32$. Note any $n>352$ with leading digit $3$ will have a digit $6$ or more and also the digit $3$. So $\frac{n}{11}\ge 45$, a contradiction. So nothing with leading digit $3$ works. Case 4: $n$ has leading digit $4$. $n\in \{407, 418, 429\}$ don't work as they each have a digit greater than or equal to $7$, but $\frac{n}{11}<49$. $n=440=40\cdot 11$ doesn't work as $4^2+4^2+0^2=32$. $n=451=41\cdot 11$ doesn't work as $4^2+5^2+1^2=42$. $n=462=42\cdot 11$ doesn't work as all the squares we are adding are multiples of $4$. All $n>462$ with leading digit $4$ have a digit greater than or equal to $7$, but $49\cdot 11>500$. So nothing with leading digit $4$ works. Case 5: $n$ has leading digit $5$. $n=506=46\cdot 11$ doesn't work as the sum of squares is odd. $n\in \{517, 528, 539\}$ don't work because they have two digits greater than or equal to $5$, and $\frac{n}{11}<50$. $n=\boxed{550}=50\cdot 11$ works. If $n>550$ has leading digit $5$, it must have another digit greater than or equal to $6$, but this is a contradiction as $\frac{n}{11}<5^2+6^2$. So there is one solution if $n$ has leading digit $5$. Case 6: $n$ has leading digit $6$. Note $n\in \{605,616,627,638,649\}$ don't work as they each have the digit $6$ and another digit greater than or equal to $5$, but they are less than $61\cdot 11=671$. Any $n>649$ with leading digit $6$ must have two digits greater than or equal to $6$. But $\frac{n}{11}<6^2+6^2$, a contradiction. So nothing with leading digit $6$ works. Case 7: $n$ has leading digit $7$. $n=704=64\cdot 11$ doesn't work as $7^2+4^2=65$. $n\in \{715, 726, 737, 748, 759\}$ don't work as the sum of squares of their digits for each of them is greater than $74$, but they are all less than $74\cdot 11$. $n>759$ with leading digit $7$ must have two digits greater than or equal to $7$, but $\frac{n}{11}<7^2+7^2$. So nothing with leading digit $7$ works. Case 8: $n$ has leading digit $8$. $n=\boxed{803}=73\cdot 11$ works. $n\in \{814, 825, 836, 847, 858, 869\}$ don't work as the sum of squares of their digits for each of them is greater than $80$, but they are all less than $80\cdot 11$. $n=880=80\cdot 11$ doesn't work as $8^2+8^2>80$. $n=891=81\cdot 11$ doesn't work as $8^2+9^2>81$. So there is one solution if $n$ has leading digit $8$. Case 9: $n$ has leading digit $9$. Note that $n$ can't have any other digit greater than or equal to $4$ because otherwise the sum of squares of digits exceeds $90$. $n=902=82\cdot 11$ doesn't work as $9^2+0^2+2^2=85$. $n=913=83\cdot 11$ doesn't work as $9^2+1^2+3^2=91$. All other $n$ with leading digit $9$ don't work as they have another digit greater than or equal to $4$. So nothing works with leading digit $9$. Thus, our only solutions are $n=550$ and $n=803$.

Let $xyz$ be such a 3-digit number $(x,y,z$ represents digits.) So $11|xyz \implies y=z+x \text{ or } y=z+x-11.$ In the first case, $x^2+z^2+(x+z)^2=10x+z.$ In the second case, $x^2+z^2+(x+z-11)^2=10(x-1)+z.$ In the first case LHS is even so $z\in \{0,2,4,6,8\}.$ In the second case LHS is odd so $z\in \{1,3,5,7,9\}.$ Analysis of the the 10 quadratic equations will lead to the solution $\boxed{n=\{550,803\}}.$

I have discussed this problem on my YouTube channel: Video link: Video Solution

Solution one: 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.

@above I would suggest in using $LaTeX.$ It's a word processor for mathematicians (and other people too.) You might learn from other sites but the most basic way is to use a dollar sign before and after your math texts. For example $ m=a^2+b^2+c^2 $ becomes $m=a^2+b^2+c^2.$

mylist=[]for a in range(9): a=a+1 for b in range(10): for c in range(10): k=100*a+10*b+c if k%11==0: if 100*a+10*b+c==11*(a**2+b**2+c**2): mylist.append(a) mylist.append(b) mylist.append(c) print(mylist) mylist=[] continue continue continue continuemylist=[] for a in range(9): a=a+1 for b in range(10): for c in range(10): k=100*a+10*b+c if k%11==0: if 100*a+10*b+c==11*(a**2+b**2+c**2): mylist.append(a) mylist.append(b) mylist.append(c) print(mylist) mylist=[] continue continue continue continueRunResetPop Out /

Case 1: digits are $a, a+b, b$ $\overline{ab} = a^2+b^2+(a+b)^2$ $10a+b = 2a^2+2b^2+2ab$ $2a^2+a(2b-10)+(2b^2-b)=0$ $a = \frac{10-2b \pm \sqrt{(2b-10)^2-8(2b^2-b)}}{4}$ Plugging in values of $b$ from $0$ to $9$, get that $b=0,a=5$ works and nothing else, which results in $550$ Case 2: digits are $a,a+b-11,b$ $10(a-1)+b = a^2+b^2+(a+b-11)^2$ $10a-10+b = a^2+b^2+(a+b-11)^2$ $2a^2+a(2b-32)+(131-23b+2b^2)=0$ $a=\frac{32-2b \pm \sqrt{(2b-32)^2-8(131-23b+2b^2)}}{4}$ Plugging in values of $b$ from $0$ to $9$, get that $b=3,a=8$ works and nothing else, which results in $803$ Total: $\boxed{550, 803}$

note, i did not copy this, i wrote this for my python class: for n in range(110,1000,11): # check each possibility # grab the digits of n hundreds = n // 100 tens = (n // 10) % 10 ones = n % 10 # check if n/11 is the sum of the squares of the digits if n // 11 == hundreds**2 + tens**2 + ones**2: print(n)for n in range(110,1000,11): # check each possibility # grab the digits of n hundreds = n // 100 tens = (n // 10) % 10 ones = n % 10 # check if n/11 is the sum of the squares of the digits if n // 11 == hundreds**2 + tens**2 + ones**2: print(n)RunResetPop Out /

since you have so much time during an imo you could just list out all three-digit numbers that are a multiple of 11 and bash