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Problem 255

Number theory Difficulty 2.1 Prove it CEMC Fryer · Canada · 2026

List PP contains all positive
integers from 11 to 2532^{53}, inclusive.

How many numbers in list PP can be written as 2k2^k where kk is a positive integer?
Since 4=224=2^2, every power of 44 can be written as a power of 22. For example, 434^3 can be written as 43=(22)3=22×3=264^3=\left(2^2\right)^3=2^{2\times 3}=2^6.
In general, $4n=(22)n=22×n=22n\$4^n=\left(2^2\right)^n=2^{2\times n}=2^{2n}.Howmanynumbersinlist. How many numbers in list Pcanbewrittenas can be written as 44^\ellwhere where $\ell\$ is a positive integer?
Determine how many numbers in list PP can be written as 4r4^r where rr is a positive integer, but cannot be
written as 8t8^t where tt is a positive integer.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

List PP contains all positive
integers from 11 to 2532^{53} inclusive, and so PP contains each integer of the form 2k2^k where kk is a positive integer from 11 to 5353 inclusive.

Thus, there are 5353 such numbers in
the list PP.
Since 4=(22)=224^\ell=(2^2)^\ell=2^{2\ell}, then 44^\ell is in list PP exactly when 222532^{2\ell}\leq2^{53} or 2532\ell\leq53 for positive integers \ell. The largest positive integer \ell for which 2532\ell\leq53 is 2626 (since 2×26=522\times26=52, and 2×27=542\times27=54).

Thus, for all positive integers $\$\ell \leq
26,, 4$4^\ell\$ is in list
PP, and so there are 2626 such numbers.

We demonstrate this relationship between the powers of 22 and 44 in the table below.

2k2^k
212^1
222^2
232^3
242^4
252^5
262^6
\cdots
2512^{51}
2522^{52}
2532^{53}

4=224^{\ell}=2^{2\ell}

41=224^1=2^2

42=244^2=2^4

43=264^3=2^6
\cdots

426=2524^{26}=2^{52}

From (b), list PP contains
the 2626 numbers 4r=22r4^r=2^{2r} for positive integers r26r\leq 26.

That is, each integer power of 44 in
PP is equal to a power of 22 whose exponent is an even positive
integer.

Since 8t=(23)t=23t8^t=(2^3)^t=2^{3t}, then each
integer power of 88 in PP is equal to a power of 22 whose exponent is a positive integer
multiple of 33.

Thus, any number in PP that is an
integer power of both 44 and 88 must be equal to a power of 22 whose positive integer exponent is both
even and a multiple of 33, and
therefore a multiple of 66.

PP contains all numbers 2k2^k for positive integers k53k\leq 53, and so PP contains the following powers of 22 whose exponent is a multiple of 66: 262^6, 2122^{12}, 2182^{18}, 2242^{24}, 2302^{30}, 2362^{36}, 2422^{42}, and 2482^{48}.

For positive integers rr and tt, there are 88 numbers in PP that can be written as both 4r4^r and as 8t8^t, and so there are 268=1826-8=18 numbers in PP which can be written as 4r4^r but cannot be written as 8t8^t.

We demonstrate this relationship between the powers of 22, 44
and 88 in the table below.

4r=22r4^r=2^{2r}
414^1
424^2
43=264^3=2^6
444^4
454^5
46=2124^6=2^{12}
\cdots
424=2484^{24}=2^{48}
4254^{25}
4264^{26}

8t=23t8^t=2^{3t}

82=268^2=2^{6}

84=2128^4=2^{12}
\cdots
816=2488^{16}=2^{48}

4r4^r
not 8t8^t
414^1
424^2

444^4
454^5

\cdots

4254^{25}
4264^{26}

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.