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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 ? ' |
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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? ' |
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How many distinct four-digit numbers are divisible by and have as their last two digits? ' |
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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? ' |
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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 ? ' |
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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 ? ' |
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The number obtained from the last two nonzero digits of is equal to . What is ? ' |
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A year is a leap year if and only if the year number is divisible by 400 (such as 2000) or is divisible by 4 but not 100 (such as 2012). The 200th anniversary of the birth of novelist Charles Dickens was celebrated on February 7, 2012, a Tuesday. On what day of the week was Dickens born? ' |
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First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities. After some inspection, it can be seen that , and , so Therefore, can be written as Keep in mind that can be , five choices; Now, examine the tens digit, by using and to find the tens digit (units digits can be disregarded because will always work) 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) and this simplifies to From inspection, when This gives you choices for , and choices for , so the answer is |
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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 |
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Solution 1The 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 Solution 2We 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 . |
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Solution 1To 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. 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. Solution 2A 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. |
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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 . |
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Solution 1When 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 . Solution 2Let equal , where through are digits. Therefore, 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. Solution 3Since 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.) NotesThe part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . 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. |
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Solution 1The 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 . Solution 2We 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 . |
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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 . |
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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 : .
Alternate SolutionNotice 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.
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We will use the fact that for any integer , First, we find that the number of factors of in is equal to . Let . The we want is therefore the last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find . If we divide by , we can write as where where every number in the form is replaced by . The number can be grouped as follows: Using the fact that ,we can deduce that . Therefore . Finally, combining with the fact that yields . |
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Each year we go back is one day back, because . Each leap year we go back is two days back, since . A leap year is GENERALLY every four years, so 200 years would have = leap years, but the problem points out that 1900 does not count as a leap year. This would mean a total of 151 regular years and 49 leap years, so = days back. Since , four days back from Tuesday would be |