**
***Title*
Blocked Off
**
***Problem Statement*
Jeff had fewer than 100 blocks. When he made five equal rows, he had one left
over; with four equal rows, he had one left over; and with nine equal rows, he
had none left over. How many blocks did Jeff have?
**
***Problem setup*
Jeff had less than 100 blocks which
equaled 5 * n +1, 4 * n + 1, and 9 * n.
*Plans to Solve/Investigate the
Problem*
At first I tried the guess and check
method to determine a number that would satisfy the first 2 requirements, but
then I remembered the third criteria. So I looked at the products of 9.
*Investigation/Exploration of the
Problem*
My first guesses were 96 and 71, but
they did not = 4 * n + 1. After I listed the multiples of 9 through 9 *
11, I noticed that 81 was divisible by both 5 and 4 with each leaving a
remainder of 1. 81 is the answer!
*Extensions of the Problem*
Suppose that all the conditions about
Jeff's blocks remain except that he has fewer than 1000 blocks. How many blocks
might Jeff have?
I tried multiplying the multiples of
9 (through 9 * 11) by 2, but that wouldn't work since if you divided the numbers
in the new list by 5 or 4, you would not get 1 as a remainder. The same if
I multiplied by 3, 4, 5, 6, 8, 10. Then I tried multiplying by 9, but,
again, only 81 satisfied all of the requirements. Then I made a spread
sheet to list all multiples of 9 until the numbers reached over 1000. 5 *
n + 1 would have either a 6 or a 1 in the ones place. I need only an odd
number in the ones place, so my answers must end in 1 to satisfy that criteria.
Multiples of 4 plus 1 would have either 1, 5, 9, 3, 7 in the ones place, so to
satisfy both 5 * n + 1 and 4 * n + 1, the answers must have a 1 in the ones
digit. After eliminating all other numbers on the spread sheet, I was left
with 81, 171, 261, 351, 441, 531, 621, 711, 801, 891, and 981. Next I
checked each to see if it was divisible by 4 with a remainder of 1. That
eliminated 171, 351, 531, 711, and 891.
Jeff could have 81, 261, 441, 621,
801, or 981 blocks.
**
***Author & Contact*
../vhughes |