Problems & Puzzles: Puzzles

 Puzzle 602. T(73)=37*73 JM Bergot made me notice that 1+2+3+4+5+...73=2701=37*73 This means that the Triangular number, T(73)=2701=37*73, is a semiprime composed by mirrored primes Naturally the first question coming to our mind is, are there other Triangular numbers alike? I found an infinite family of solutions: Every Triangular number of the type: T[7(9)x3]=[7(9)x3]*[3(9)x7], for x=0 to infinite. But I was unable to find an 0

Contributions came from Hakan Summakoğlu, J.K. Andersen.

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Hakan wrote:

Q1: I found 2 triangular numbers and an infinite triangular family formed by multiplying of mirrored numbers.

1+2+3+4+…. +376=116*611=70876
1+2+3+4+..... +845=507*705=357435

Other infinite family: T[11(9)x6]=[7(9)x8]*[8(9)x7], for x=0 to infinite.

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J.K. Andersen wrote:

Q2. There are no solutions up to 150000 digits.
Heuristics say there are probably no solutions at all.

A099190: Numbers n such that 8*10^n-7 is prime.
1, 3, 5, 6, 10, 12, 13, 39, 48, 54, 64, 82, 147, 148, 360, 399, 1638,
1876, 2146, 2194, 15789, 23074, 38466, 68400

searched up to n<=100000. The only of these for which 4*10^n-3 is also
prime is n=1, giving the known T(73)=37*73.

A101398: Numbers n such that 4*10^n-3 is prime.
1, 2, 14, 20, 30, 44, 66, 260, 872, 8846, 26744, 57506

searched up to n<=60000.

Some of the OEIS numbers are only prp's but it's irrelevant for us
when the other number is known to be composite.

I used srsieve and PrimeForm/GW to test 8*10^n-7 for n = 100000 to
150000 when none of the two numbers had a prime factor below 10^10.
No prp's were found.

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Emmanuel Vantieghem wrote (Set 2011)

It is elementary (but a littlebit difficult to write down as you will see in a moment) to prove that, when

1+2+...+k = k(k+1)/2 = kR(k),  or  2R(k)-1 = k,

we have that  k  is of the form 399...97.

I'll give only an outline of a proof (details leaving to the reader).

We assume  k > 3 (proof for smaller  k  eventually leaving to the PC).

Let  a(0), a(1), ..., a(n)  be the decimal digits of  k  (from left to right).

Then : R(k) = a(n)10^n+a(n-1)10^(n-1)+...+a(1)10+a(0)

= a(n)10^n+10x+a(0), with  x = a(n-1)10^(n-2)+...+a(2)10+a(1).

Assume we are given  2R(k)-1 = k = a(0)10^n+10R(x)+a(n).  This implies  2a(0)-1 == a(n) (mod 10)  and

either     2a(n) = a(0)  (*)      or  2a(n)+1 = a(0) (**)  (elemantary arithmetic).

From (*) follows  a(n) = 7, which is impossible since  2R(k)-1  has as many digits as  R(k).

Thus, (**) must hold, or : a(n) = 3  and hence  a(0) = 7.  From this it follows :

20x+10 = 10^n+10R(x)   or :    2x+1 = 10^(n-1)+R(x).      (1)

Looking at the number of digits of  x, we see that the left digit of  x  must be >= 5.

We have either     2a(n-1) = 10+a(1) (***)    or     2a(n-1)+1 = 10+a(1)  (****).

From (***) we deduce together with  2a(n-1)+1 == a(1) mod 10  that  a(n-1)  = 3, impossible.

Hence, (****) holds and thus  a(n-1) = a(1) = 9.

When we write  x  as  9(10^(n-2))+10y+9, we can deduce from (1) :

2(9(10^(n-2))+10y+9)+1=10^(n-1)+9(10^(n-2))+10R(y)+9   or :

18(10^(n-2))+20y+19 = 19(10^(n-2)+10R(y)+9  or :  2y+1 = 10^(n-3)+R(y).

This equation is of the same kind as  (1)  and will lead analoguously to  a(n-2) = 9 = a(2).

Using suitable induction will finally give QED.

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