Curious Identities Involving Integer Products
This is a continuation of the page where I started to collect identities with expressions that include repeated identical digits. To make describing such numbers a manageable undertaking, I'll use subscripts to denote the number of repetitions of a digit, e.g., 1_{5}3_{2} will denote 1111133, i.e., five 1s followed by 2 3s. In general d_{n} means a sequence of n d's.
The identities evolved from CSIRO newsLetter, MathByEmail. An early morning exchange on twitter.com with Gary Davis and subsequent comments by James Tanton, who has observed that the repunits (the integers 111..11) with an even number of 1s are represented by the difference of squares of which one is in the form 555...56 in two ways.
3_{k} · 3_{k1}4 = 1_{k}2_{k}, e.g., 3333 · 3334=11112222. (Proof)
3_{k} · 3_{k1}5 = 1_{k}5_{k}, e.g., 3333 · 3335=11115555. (Follows from the previous one)
3_{k} · 3_{k1}6 = 1_{k}8_{k}, e.g., 3333 · 3336=11118888. (Follows from the previous one)
6_{k} · 6_{k1}7 = 4_{k}2_{k}, e.g., 6666 · 6667 = 44442222. (Proof)
6_{k} · 6_{k1}8 = 4_{k}8_{k}, e.g., 6666 · 6668 = 44448888. (Follows from the previous one)
9_{k} · 10_{k1}1 = 9_{2k}, e.g., 9999 · 10001 = 99999999.
5_{k1}6²  4_{k1}5² = 1_{2k}, e.g., 5556²  4445² = 11111111. (Proof)
5_{k1}6²  5_{k}² = 1_{k+1}, e.g., 5556²  5555² = 11111.
6_{k} · 6_{k1}9 = 4_{k1}5_{k}4, e.g., 6666 · 6669 = 44455554.
Subscripts can be easily converted to decimal notations. Observe that
1_{t} = 9_{t} / 9 = (10^{t}  1) / 9.
So, for example,
3_{k}5_{m}6_{n}7 = 7 + 10 · 6 · (10^{n}  1) / 9 + 10^{n + 1} · 5 · (10^{m}  1) / 9 + 10^{m + n + 1} · 3 · (10^{k}  1) / 9.
Number Curiosities
 Number 8 Is Interesting
 Curious Identities Involving Integer Squares
 Curious Identities Involving Integer Products
 Decimal Sums of Successive Integers
 Curious Identities In Pythagorean Triangles
 Hardy's Example of NonSerious Theorems
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Copyright © 19962018 Alexander Bogomolny

3_{k} · 3_{k1}4 = 1_{k}2_{k}.
3_{k} · 3_{k1}4  = 3 · (10^{k}  1)/9 × [3 · (10^{k}  1)/9 + 1] 
= (10^{k}  1)/3 × [(10^{k}  1)/3 + 1]  
= (10^{2k}  2 · 10^{k} + 1)/9 + (10^{k}  1)/3  
= (10^{2k} + 10^{k}  2)/9  
= (10^{2k}  1)/9 + (10^{k}  1)/9  
= 1_{2k} + 1_{k}  
= 1_{k}2_{k}. 
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Copyright © 19962018 Alexander Bogomolny

6_{k} · 6_{k1}7 = 4_{k}2_{k}.
6_{k} · 6_{k1}7  = 6(10^{k}  1)/9 × (6(10^{k}  1)/9 + 1) 
= 2(10^{k}  1)/3 × (2(10^{k}  1)/3 + 1)  
= 4(10^{2k}  2 · 10^{k} + 1)/9 + 2(10^{k}  1)/3  
= 4(10^{2k}  2 · 10^{k} + 1)/9 + 6(10^{k}  1)/9  
= 4 · 10^{2k}  2 · 10^{k}  2)/9  
= 4 · (10^{2k}  1)/9  2 · (10^{k}  1)/9  
= 4_{2k}  2_{k}  
= 4_{k}2_{k} 
Of course it is alos possible to derive this identity from
6_{k} · 6_{k1}8 = 4_{k}8_{k},
which is the next identity. Subtracting 6_{k} from both sides gives
6_{k} · 6_{k1}7 = 4_{k}2_{k},
To remind:
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