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SOURCE:COMPETITION Number of Problems: 12. FOR PRINT ::: (Book)
Bernardo chooses a three-digit positive integer and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer . For example, if , Bernardo writes the numbers and , and LeRoy obtains the sum . For how many choices of are the two rightmost digits of , in order, the same as those of ?
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
Say that
also that
After some inspection, it can be seen that , and , so , , ,
Therefore, can be written as and can be written as
Keep in mind that can be , five choices; Also, we have already found which digits of will add up into the units digits of .
Now, examine the tens digit, by using and to find the tens digit (units digits can be disregarded because will always work) Then we see that and take it and to find the last two digits in the base and representation. Both of those must add up to
()
Now, since will always work if works, then we can treat as a units digit instead of a tens digit in the respective bases and decrease the mods so that is now the units digit.
Say that (m is between 0-6, n is 0-4 because of constraints on x) Then
and this simplifies to
From inspection, when
This gives you choices for , and choices for , so the answer is
In base , the number ends in the digit . In base , on the other hand, the same number is written as and ends in the digit . For how many positive integers does the base--representation of end in the digit ?
We want the integers such that is a factor of . Since , it has factors. Since cannot equal or , as these cannot have the digit 3 in their base representations, our answer is
Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82, and 91. What was the last score Mrs. Walters entered?
The first number is divisible by 1.
The sum of the first two numbers is even.
The sum of the first three numbers is divisible by 3.
The sum of the first four numbers is divisible by 4.
The sum of the first five numbers is 400.
Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.
Case 1: 76 is the last number entered.
Since , the fourth number must be divisible by 3, but none of the scores are divisible by 3.
Case 2: 80 is the last number entered.
Since , the fourth number must be . That number is 71 and only 71. The next number must be 91, since the sum of the first two numbers is even. So the only arrangement of the scores
We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers , which gives 2,1,2,1,1, respectively. Clearly the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is .
How many distinct four-digit numbers are divisible by and have as their last two digits?
To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are , the sum of the digits is (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.
,
, and so on.
However since the largest four-digit number ending with is , the maximum sum is
.
Using that process we can fairly quickly compile a list of the sum of the first two digits of the number.
Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers in separate cases.
And finally, we add the number of elements in each set.
A number divisible by has all its digits add to a multiple of The last two digits are and and add up to Therefore the first two digits must add up to digits (including ) are are and are The following combinations are equivalent to :
Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.
The first term of a sequence is . Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the term of the sequence?
Performing this operation several times yields the results of for the second term, for the third term, and for the fourth term. The sum of the cubes of the digits of equal , a complete cycle. The cycle is... excluding the first term, the , , and terms will equal , , and , following the fourth term. Any term number that is equivalent to will produce a result of . It just so happens that , which leads us to the answer of .
What is the units digit of ?
Since :
Therefore, the units digit is
Let be a -digit number, and let and be the quotient and the remainder, respectively, when is divided by . For how many values of is divisible by ?
When a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, .
Therefore, can be any integer from to inclusive, and can be any integer from to inclusive.
For each of the possible values of , there are at least possible values of such that .
Since there is "extra" possible value of that is congruent to , each of the values of that are congruent to have more possible value of such that .
Therefore, the number of possible values of such that is .
Let equal , where through are digits. Therefore,
We now take :
The divisor trick for 11 is as follows:
"Let be an digit integer. If is divisible by , then is also divisible by ."
Therefore, the five digit number is divisible by 11. The 5-digit multiples of 11 range from to . There are divisors of 11 between those inclusive.
Since is a quotient and is a remainder when is divided by . So we have . Since we are counting choices where is divisible by , we have for some . This means that is the sum of two multiples of and would thus itself be a divisor of . Then we can count all the four digit divisors of as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)
The part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . For a digit number we get , as claimed.
Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a , the result will have the same remainder modulo as the original number.
What is the remainder when is divided by 8?
The sum of any four consecutive powers of 3 is divisible by and hence is divisible by 8. Therefore
is divisible by 8. So the required remainder is . The answer is .
We have . Hence for any we have , and then .
Therefore our sum gives the same remainder modulo as . There are terms in the sum, hence there are pairs , and thus the sum is .
Let . What is the units digit of ?
So, . Since is a multiple of four and the units digit of powers of two repeat in cycles of four, .
Therefore, . So the units digit is .
For , let , where there are zeros between the and the . Let be the number of factors of in the prime factorization of . What is the maximum value of ?
The number can be written as .
For we have . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have .
For we have . For the value in the parentheses is odd, hence .
This leaves the case . We have . The value is obviously even. And as , we have , and therefore . Hence the largest power of that divides is , and this gives us the desired maximum of the function : .
Notice that 2 is a prime factor of an integer if and only if is even. Therefore, given any sufficiently high positive integral value of , dividing by yields a terminal digit of zero, and dividing by 2 again leaves us with where is an odd integer. Observe then that must be the maximum value for because whatever value we choose for , must be less than or equal to 7.