31

Hardy was revisiting Ramanujan and this time arrived in taxicab number 1792. He remarked that this number seemed rather dull.

Ramanujan replied, "It is very interesting, it is the product of the first six digits of a trillionth power."

Please find a whole number such that, when raised to the power 1 000 000 000 000, its first six digits (from left to right) multiply to give 1792.

Tom
  • 70,468
  • 4
  • 249
  • 361
  • 7
    I think Ramanujan would've called that number "billion". – msh210 Dec 29 '21 at 16:45
  • 3
    Even I would have once have called this number a billion. It's safer for this question to have a higher power and a lakh crore should do. – Tom Dec 29 '21 at 17:00
  • 3
    This isn't a very interesting property. You can pick any sequence of 6 digits where the first digit isn't 0 and have those 6 digits be the first 6 digits of a trillionth power, so any product of 6 single-digit integers is the product of the first 6 digits of a trillionth power. – user2357112 Dec 30 '21 at 06:02
  • @Tom Is the accepted solution the intended solution? Or did you have something else in mind? – Steven Gubkin Dec 30 '21 at 14:31
  • @Steven - the solution is as intended. I should say it's (adapted) from "Find a positive integer $n<10,000,000$ such that the first four digits (in the decimal expansion) of $n^{1,000,000}$ are all different" from Richard Stanley's page – Tom Dec 30 '21 at 16:39
  • I'm not sure that the story as told above is true. I remember (not sure from where) that Ramanujan's answer was 'it is the smallest number expessible as the sum of two positive cubes in two ways'. Hardy then asked what the corresponding number for fourth powers would be, and Ramanujan answered after a short pause that he didn't know, but that it would be very large. See https://vedicmathschool.org/story-of-hardy-ramanujan-number-1729/ – Peter Dec 31 '21 at 18:37
  • 3
    @Peter: The story is completely made up for the puzzle. Note how the number is 1792, not 1729. It's presented as a sequel to the story of the actual Hardy-Ramanujan number. – user2357112 Jan 01 '22 at 21:54
  • @user2357112supportsMonica Oops! Sorry! – Peter Jan 03 '22 at 10:49

2 Answers2

34

First observe that the first $6$ digits of

$e = \mathbf{2.71828}\ 18284 \ldots$

happen to multiply to $1792$. Now all we need is to remember the bounds

$(1+\frac 1 n)^n < e < (1+\frac 1 n)^{n+1}$ (for positive $n$)

to find that the number

$1\ 000\ 000\ 000\ 001$

does the trick.

Indeed, setting $n = 1\ 000\ 000\ 000\ 000$ we get

$e > 1.000\ 000\ 000\ 001^{1\ 000\ 000\ 000\ 000} > \frac e {1.000\ 000\ 000\ 001}$.

Multiplying through with $1\ 000\ 000\ 000\ 000^{1\ 000\ 000\ 000\ 000}$ we confirm that the first $10$ or so digits of the $1\ 000\ 000\ 000\ 000$th power and of

$e$

are the same as desired.

loopy walt
  • 21,184
  • 33
  • 95
5

The smallest natural number that fits the bill is

1931

but

that's 11 years after his death

so I guess the other answer is better...

Code to find that number, computing not the exact powers but only their first 100 digits but computing lower and upper bound and checking that their first six digits are the same so we know they're exact:

import math

i = 0
while True:
    b = B = i
    e = 1012
    p = P = 1
    while e:
        if e % 2:
            p = b
            P = B
            while p > 10100:
                p //= 10
                P = -(P // -10)
        e //= 2
        b = b
        B = B
        while b > 10**100:
            b //= 10
            B = -(B // -10)
    p = str(p)
    P = str(P)
    assert len(p) == len(P)
    assert p[:6] == P[:6]
    if math.prod(map(int, p[:6])) == 1792:
        print(i)
        break
    i += 1

Attempt This Online!

Kelly Bundy
  • 251
  • 1
  • 6