Olympiad Maths Prep

Track / Stage 4 / 167 of 340 #427 of 2000

Problem 427

AMC 12 late, AIME early
Number theory Difficulty 4.8 Find the answer

6. Find the smallest natural number ending in the digit 2 that doubles when this digit is moved to the beginning.

Official solution

Answer: 105263157894736842

Solution: Let's write the number in the form *... 2 and gradually restore the "asterisks" by multiplying by 2:

...2×2=..4* * * . . . * * 2 \times 2=* * * . . * * 4

...42×2=...84* * * . . . * 42 \times 2=* * * . . . * 84

842×2=684* * * \ldots * 842 \times 2=* * * \ldots * 684

..6842×2=3684* * * . . * 6842 \times 2=* * * \ldots * 3684

***...*36842 2=736842=* * * \ldots * 73684

***... 736842×2=...473684* 736842 \times 2=* * * . . . * 473684

***...*4736842 x 2=..94736842={ }^{* * *} \ldots . . * 9473684

...

105263157894736842×2=2105263157105263157894736842 \times 2=2105263157

89473684

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.