Expert: Scott A Wilson - 10/27/2008

Question
QUESTION: Hey - I'm prepping for the GREs and it's making me go back to high school problem solving that I can't remember. Worse, no calculators allowed! Could you explain this one to me?

How many 3-digit positive integers are odd and do not contain the digit '5'?

THANKS!

ANSWER: There are 10 choices for each digit under normal circumstances, so there are 10*10*10=1,000 choices for three digit numbers (you know, 0-999).

If they are odd, only half of these are odd, so it's 500 (10*10*5).

If they contain no 5, there are only 9 ways of choosing the first number and 9 ways of choosing the second number.  Since 5 is odd, there are only 4 ways of choosing the last number.

The answer is 9*9*4










=324.

---------- FOLLOW-UP ----------

QUESTION: Hi again, Thanks for your help. Unfortunately,(9)(9)(4)=324.
The answer given is 288.

Your method makes sense to me except that I think you accounted for the 3rd digit twice. It can't be odd so you eliminated that when you divided by two but then you also eliminated the possibility of a 5. Doesn't that compensate for the 3rd digit not being a 5, twice? Unfortunately again, (9)(9)(5) is not the answer either so that can't be the explanation.

If you have any ideas, I'd love to hear them. THANKS

Answer
I checked in a spreadsheet and put down 000 through 999.
If any of the digits were 5, I tossed them out.
If the last digits was odd, I tossed it out.
I got 324.

I also put in the fact that all of the numbers must be different, but then I only obtained 224.

I then noticed if they only give 9 choices to the first digit, then make the second one different with only 8 choices and still 4 choices for the last one, they get 288.

So, the way they problem is read, there are 324.
If no two numbers are allowed to be the same, there are 224.
If the first two digits are required to be different, there are 224.

I would tell the teacher about this.

001,003,007,009,011,013,017,019,021,023,027,029,031,033,037,
039,041,043,047,049,061,063,067,069,071,073,077,079,081,083,
087,089,091,093,097,099,101,103,107,109,111,113,117,119,121,
123,127,129,131,133,137,139,141,143,147,149,161,163,167,169,
171,173,177,179,181,183,187,189,191,193,197,199,201,203,207,
209,211,213,217,219,221,223,227,229,231,233,237,239,241,243,
247,249,261,263,267,269,271,273,277,279,281,283,287,289,291,
293,297,299,301,303,307,309,311,313,317,319,321,323,327,329,
331,333,337,339,341,343,347,349,361,363,367,369,371,373,377,
379,381,383,387,389,391,393,397,399,401,403,407,409,411,413,
417,419,421,423,427,429,431,433,437,439,441,443,447,449,461,
463,467,469,471,473,477,479,481,483,487,489,491,493,497,499,
601,603,607,609,611,613,617,619,621,623,627,629,631,633,637,
639,641,643,647,649,661,663,667,669,671,673,677,679,681,683,
687,689,691,693,697,699,701,703,707,709,711,713,717,719,721,
723,727,729,731,733,737,739,741,743,747,749,761,763,767,769,
771,773,777,779,781,783,787,789,791,793,797,799,801,803,807,
809,811,813,817,819,821,823,827,829,831,833,837,839,841,843,
847,849,861,863,867,869,871,873,877,879,881,883,887,889,891,
893,897,899,901,903,907,909,911,913,917,919,921,923,927,929,
931,933,937,939,941,943,947,949,961,963,967,969,971,973,977,
979,981,983,987,989,991,993,997,999.

Now that's 21 rows of 15 numbers with 9 in the last row, which is 21*15 + 9 = 315 + 9 = 324.

If you need another list, I can provide them as well.

Maybe there was something special in the problem that wasn't include.