math questions....show your workkk?

Question

a. gerry wants to mail a package that requires $1.53 in postage. if he has only 5- cent and 8-cent stamps, what is the smallest number od stamps he could use that would total exactly $1.53?

b. let Pn = 1^n + 2^n + 3^n + 4^n. find the number of integers n for which 1 _< n _< 100 and Pn is a multiple of 5.

note. those _< are the equal to or less than signs.

Chris L · Accepted Answer

There are algebraic ways to solve these, but they're easy enough without algebra.

a.
The more 8-cent stamps we have, the less stamps we will use in total.  The remainder will have to be paid in 5-cent stamps; multiples of 5 always end in 0 or 5.  So the total paid from our 8-cent stamps will have to end in either 3 (3 - 0) or 8 (13 - 5).

So we're looking for the largest multiple of 8, that is less than 153, that ends in 3 or 8 (actually, just 8, because multiples of 8 are always even).  This is 128, which is 8 * 16.  The remainder is 25 cents, which is 5 * 5.  So we have 16 8-cent stamps and 5 nickel stamps to make a total of 21 stamps.

b.
To know whether or not Pn is a multiple of 5, we only have to know the last digit, and whether it's a 5 or 0.  Conveniently, the last digits of powers follow patterns.  Let's see what they are:
1+2+3+4 <-- these are all those numbers to the 1st power

1+4+9+16 <-- these are all of them to the 2nd power
1+4+9+6 <-- just the last digits

1+8+27+64 <-- these are all of them to the 3rd power
1+8+7+4 <-- just the last digits
At this point, notice that if you multiplied the "6" from the last digits of the 2nd power list by 4 again, you'd get 24 instead of 64 here.  The last digit, though, is still 4.  In other words, we don't need to keep multiplying these huge numbers.  We can just multiply the last-digit row to get our next set.

1+16+21+16 <-- I got these by multiplying the 3rd power last-digit row by 1, 2, 3, 4
1+6+1+6 <-- so these are the last digits for the 4th power
1+16+81+256 <-- see, they actually are the last digits of these!

1+12+3+24 <-- multiplying 4th power last-digit row by 1, 2, 3, 4
1+2+3+4 <-- last digits for 5th power
Hey wait!  We've come back to 1+2+3+4 again!

So, our pattern is:
1+2+3+4 = 10
1+4+9+6 = 20
1+8+7+4 = 20
1+6+1+6 = 14
And then it repeats.  Since the first three will all add up to a number that ends in '0', this means that 3 out of every 4 integers "n" will make Pn a multiple of 5.

So, out of a set of 1 to 100, 75%, or 75 of them will work.

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math questions....show your workkk?

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