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Four equal digits (Posted on 2021-07-20) Difficulty: 3 of 5
Find all possible integer values of n such that 12n2 + 12n + 11 is a 4-digit number with equal digits.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Analytical Puzzle Solution Comment 5 of 5 |
We observe that:
12(n+1)^2=12n^2+24n+12>12n^2+12n+11>=1111
=> 12(n+1)^2 > 1111>1104
=> (n+1)^2 > 92> 81
=> n+1>9
=> n>8
=>n>=9
Also, we have:
12*n^2< 12n^2+12n+11<=9999
=> n^2< floor(9999/12) =833 <841
=> n<29
=> n<= 28
Hence, 9<=n<=28.......(i)
Now, any 4-digit number with equal digits has the form pppp, where p is a positive digit between 1 and 9 inclusively.
Accordingly,  denoting the required 4-digit number by M, we have:
M=1111*p=11*101*p
=> M=0(mod 11)
Accordingly,  by the given conditions, wemust have:
12n^2+12n+11=0(mod 11)
=> n^2+n=0(mod 11)
=> n(n+1)=0(mod 11)
As, gcd(n,n+1)=1, it follows that:
EITHER: n=0(mod 11)  |               (ii)
OR: n+1=0(mod 11)     |

Let f(n) = 12n^2+12n+1=12n(n+1)+11
Recalling from (i), that 9<=n<=28, we observe from (ii) that the only valid values of n are 11,12,21 and 22.
Now, using a hand-calculator, we have:
f(11)=1595 -> NOT a number formed by equal digits
f(12)=1883-> NOT a number formed NY 4 equal digits
f(21)=5555-> A NUMBER FORMED BY 4 EQUAL DIGITS
f(22)=6083,-> NOT a number formed by 4 equal digits.
From the above, we can easily conclude that the ONLY NUMBER satisfying all the conditions of the given problem is 5555.

Accordingly, , the  possible positive integer value satisfying all the conditions of the puzzle under reference is 5555.
Therefore, we have:
12n^2+12n+11=5555
or, 12n^2+12n = 5544
or, n^2+n = 462
or, n^2+n-462 =0
or, (n+22)(n-21)=0
or, n=-22, 21
Consequently,  the required possible integer values of n are -22, and 21.  

Edited on December 27, 2022, 12:50 am
  Posted by K Sengupta on 2022-01-07 01:15:23

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