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## Lotto Luck

 Category: Logic Submitted By: cnmne Fun: (2.23) Difficulty: (3.09)

Professor Abacus is purchasing a ticket for the Deca Lotto. The lotto number has ten digits, using the numerals 0 through 9, each numeral used once. The clerk asked what number he wanted to pick. Professor Abacus handed the clerk a piece of paper with nine statements, saying "If you can correctly figure out the number, I will give you half of whatever I win." What is the number?

1) The sum of the first five digits is a prime number.
2) The sum of the last five digits is a triangle number*.
3) The sum of the digits in the odd positions is an odd number.
4) The sum of the middle two digits is a square number.
5) The sum of the middle four digits is a cube number.
6) The difference between the 1st and 10th digits is two.
7) The difference between the 2nd and 9th digits is three.
8) The difference between the 3rd and 8th digits is four.
9) The numeral 4 is somewhere in the first five positions.

* You can form a triangle arrangement by building it in the pattern row 1 = 1, row 2 = 2, row 3 = 3 etc. eg. 10 is a classic triangle number as per ten pin bowling. They are arranged in a triangle 1, 2, 3, 4.

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Comments on this teaser

 Posted by I_am_the_Omega 11/23/04 First, I'd like that say while it's good you explained what a triangle number is, you did the worst possible job explaining what it was. I hadn't a clue what you meant. I had to look it up. The site i looked at explained it in such a way that I got it instantly. Second, I can't say for sure, but are you positive there aren't ... oh, say, 4.096billion possible answers? I mean, seriously. lets break down your information.. sum of 1-5 is prime. Uh, ok. 1+2+3+4+7 works. So's 1+5+4+7+0. And 8+2+3+6+4. And 1+2+4+6+0. And 1+4+3+0+9. And so on... those are only for two primes.. and not even covered. Obiviously, that's 0-9. Cross out that information! Ok... lets see. The sum of the two middle is a square number? Great! Except, this helps not at all. Not at ALL, in ANY way. Why not? The number could be any number, 0-9. Why? Uh, let's think. 9 is the square number (obviously). 0+9=9. 1+8=9. 2+7=9. 3+6=9. 4+5=9. And that covers all 10 of those numbers. Same deal with the cube number bs. Difference between 1 and 10? Hmm. 9-7=2. 8-6, 7-5, 6-4, 4-2, 3-1, and 2-0 all equal 2. Cross that out. Same deal for infromation in #7 and 8. Ok.. lets see. Same thing for #3. Please explain why there's only one answer? Posted by Jimbo 11/24/04 each numeral may only be used once! Posted by javaguru 12/30/08 This was a very nice teaser. The problem actually solves fairly elegantly. The digits 0 to 9 sum to 45. The only sum of a prime and triangular number that totals 45 is 17 and 28. Next the possible choices for the square are 1, 4, 9 and 16; for the cube are 8 and 27. The cube must be from 1 to 17 greater than the square giving the possible combinations (1,8 ), (4,8 ) and (16,27). 1 can be made only from 0 and 1. This leaves either (3,4) or (2,5) to get to 8. 4 can be made from either (1,3) or (0,4) which can be combined with either (1,3) or (0,4). 16 can only be made with 7 and 9. This leaves either (3,8 ) or (5,6) to get to 27. So there are only 6 possible combinations of numbers for the center 4 digits: (0,1)+(3,4) (0,1)+(2,5) (0,4)+(1,3) (1,3)+(0,4) (7,9)+(3,8 ) (7,9)+(5,6) Three of these combinations use the same digits (0,1,3,4), so there are only 4 possible combinations of digits for the center and for the remaining places: (0,1,3,4)(2,5,6,7,8,9) (0,1,2,5)(3,4,6,7,8,9) (3,7,8,9)(0,1,2,4,5,6) (5,6,7,9)(0,1,2,3,4,8 ) We know the first 5 digits must total 17 and include the 4, so the other four digits in the first half must total 13. There are only four combinations of 4 of the remaining digits that total 13, so the first 5 digits must be 4 plus one of: 0+1+3+9=13 0+1+5+7=13 0+2+3+8=13 1+2+3+7=13 The center combinations: (0,4)+(1,3) (1,3)+(0,4) pair a 0 with the four, which would eliminate the first three totals (the zero is in the lower 5 digits) and the last one pairs a 1 with a 3, so this can't be the choice for the four inner digits. (0,1)+(3,4) eliminates all the upper digit combinations that include a 3 or both a 0 and 1, so this can't be the choice either. (0,1)+(2,5) eliminates all the upper digit combinations that include either 0 and 1 or 2 and 5, leaving only (0+2+3+8 ) and (1+2+3+7). (7,9)+(3,8 ) eliminates (0+2+3+8 ) with the 3,8 and it also eliminates (0+1+5+7) which doesn't include either a 3 or an 8, which leaves only (0+1+3+9) and (1+2+3+7). (7,9)+(5,6) eliminates all of the upper digit combinations except (0+1+5+7). So the five possible combinations for the (center two)+(next two center)+(upper 5)+(lower 5)=(upper 3)+(lower 3) are: (0,1)+(2,5)+(0,2,3,4,8 )+(1,5,6,7,9) = (3,4,8 )+(6,7,9) (0,1)+(2,5)+(1,2,3,4,7)+(0,5,6,8,9) = (3,4,7)+(6,8,9) (7,9)+(3,8 )+ (0,1,3,4,9)+(2,5,6,7,8 ) = (0,1,4)+(2,5,6) (7,9)+(3,8 )+ (1,2,3,4,7)+(0,5,6,8,9) = (1,2,4)+(0,5,6) (7,9)+(5,6)+ (0,1,4,5,7)+(2,3,6,8,9) = (0,1,4)+(2,3,8 ) Now find the combination for (upper 3)+(lower 3)=(differ by 4)+(differ by 3)+(differ by 2) conditions. (3,4,8 )+(6,7,9) = (3,7)+(---)+(4,6) (3,4,7)+(6,8,9) = (4,8 )+(3,6)+(7,9) (0,1,4)+(2,5,6) = (1,5)+(---)+(4,6) (1,2,4)+(0,5,6) = (1,5)+(---)+(4,6) = (2,6)+(---)+(---) (0,1,4)+(2,3,8 ) = (4,8 )+(3,0)+(---) Only one combination can satisfy the difference condition, so the upper/lower 3 digits must (3,4,7)+(6,8,9) = (4,8 )+(3,6)+(7,9). Condition (3) requiring the sum of the odd digits to be odd is not needed to produce the unique result: 7 3 4 2 1 0 5 8 6 9 8) Posted by opqpop 01/30/10 I did it another way but my solution was also really tedious. I didn't enjoy this problem... The possible squares are 1 4 9 16 since max sum of 2 different digits is 9+8 = 17. The possible cubes are 1 8 and 27 since max sum of 4 different digits if less than 40. Then you figure out which (square, cube) pairs are possible and get (1, 8), (4, 8), (16, 27). Let OE = odd + even number, EE = even + even number, and OO = odd + odd number. You note (1,8) is (OE, OE), (4,8) is (OO, EE) or (EE, OO), and (16, 27) is (OO, EE). From the difference clues, you know 1st and 10th digit is OO/EE. 2nd and 9th is OE. 3rd and 8th is OO/EE. The first 5 digits require 1, 3, or 5 odd numbers since it's prime. Same with every digit in an odd position since the clue says their sum is odd. You see that when you plug in OO EE or EE OO for the 5th,6th and 4th,7th digits, the parities will not satisfy the clues. Thus you know the square,cube pair must be (1,8). From there I just did guess and check. You know triangular number must be 28 and prime must be 17 since the digits add up to 45. You know 5th/6th digit is (0,1). 4th,7th digit is (2,5) or (3,4). Trying (2,5) will give you 1st/10th digit as (7,9), 2nd/9th as (3,6), 3rd/8th as (8,4). Stick the 4 on the left and 8 on the right. See if you can get 28 on the right with the rest of the number combinations. Do the same thing with (3,4). When you try (2,5) you see that 0 5 8 6 9 on the right adds up to 28. Double check with the left side 7 3 4 2 1 and see that it adds up to 17. Posted by opqpop 01/30/10 I'm repeating my previous post without the stupid 8) faces. I did it another way but my solution was also really tedious. I didn't enjoy this problem... The possible squares are 1 4 9 16 since max sum of 2 different digits is 9+8 = 17. The possible cubes are 1 8 and 27 since max sum of 4 different digits if less than 40. Then you figure out which (square, cube) pairs are possible and get (1, 8 ), (4, 8 ), (16, 27). Let OE = odd + even number, EE = even + even number, and OO = odd + odd number. You note (1, 8 ) is (OE, OE), (4, 8 ) is (OO, EE) or (EE, OO), and (16, 27) is (OO, EE). From the difference clues, you know 1st and 10th digit is OO/EE. 2nd and 9th is OE. 3rd and 8th is OO/EE. The first 5 digits require 1, 3, or 5 odd numbers since it's prime. Same with every digit in an odd position since the clue says their sum is odd. You see that when you plug in OO EE or EE OO for the 5th,6th and 4th,7th digits, the parities will not satisfy the clues. Thus you know the square,cube pair must be (1, 8 ). From there I just did guess and check. You know triangular number must be 28 and prime must be 17 since the digits add up to 45. You know 5th/6th digit is (0,1). 4th,7th digit is (2,5) or (3,4). Trying (2,5) will give you 1st/10th digit as (7,9), 2nd/9th as (3,6), 3rd/8th as (8,4). Stick the 4 on the left and 8 on the right. See if you can get 28 on the right with the rest of the number combinations. Do the same thing with (3,4). When you try (2,5) you see that 0 5 8 6 9 on the right adds up to 28. Double check with the left side 7 3 4 2 1 and see that it adds up to 17. Posted by Holografik 04/28/12 This was really frustrating me for the last couple of hours.. I didn't solve it but I think I could if I had spent an hour more. But I knew I was looking at it the wrong way and decided it wasn't worth spending so much time. I liked javagurus solution. That approach saves a lot of time.

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