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 1998 (Posted on 2013-12-05)
(1)Is there an integer N such that 1998*N = 22222.......22222 (only the digit 2 in the expression of this number)? If so, how many digits are in N?

(2)A pocket calculator is broken. It is only possible to use the function keys: + , - , =, 1/x(inverse function). All number keys and the memory funtion work. How can we calculate the product 37 * 54? (The result is obviously 1998.)

(3)Is it true that 11111^99999 + 99999^88888 is divisible by 1998? Same question with 111111^999999 + 999999^888888 ?

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 part 3 using UBASIC | Comment 3 of 5 |

The respective answers to part 3 are no and yes:

?modpow(11111,99999,1998)+modpow(99999,88888,1998)
3752
OK

3752 is not a multiple of 1998

?modpow(111111,999999,1998)+modpow(999999,888888,1998)
1998

but 1998 of course is.

 Posted by Charlie on 2013-12-05 15:46:56

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