Lesson Math circle level problems on divisibility numbers
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<H2>Math circle level problems on divisibility numbers</H2> <H3>Problem 1</H3>Find the number of 7-digit positive integers, where the sum of the digits is divisible by 3. <B>Solution</B> <pre> First 7-digit positive integer number is 1,000,000 (one million). The last 7-digit positive integer number is 9,999,999. The number of all 7-digit positive integer number is 9,999,999 - 999,999 = 9,000,000. The 7-digit positive integer numbers, whose sum of digits is divisible by 3 are those and only whose of this set that are divisible by 3. The number of such numbers is one third of 9,000,000, i.e. 3,000,000. <U>ANSWER</U>. The number of 7-digit positive integers, where the sum of the digits is divisible by 3 is 3,000,000. </pre> <H3>Problem 2</H3>Find the number of 7-digit numbers, where the sum of the digits is divisible by 9. <B>Solution</B> The solution is in 3 steps. <pre> (1) The 7-digit numbers are from 1,000,000 to 9,999,999 inclusive. The number of 7-digit numbers is 9,999,999 - 999,999 = 9,000,000. (2) In the base 10, the sum of the digits is divisible by 9 if and only if the number itself is divisible by 9. (3) The consequence is that of 9,000,000 7-digit numbers from 1,000,000 to 9,999,999 every 9-th number is divisible by 9 - hence, the number of all such numbers, where the sum of digits is divisible by 9 is 9,000,000 : 9 = 1,000,000. <U>ANSWER</U> </pre> <H3>Problem 3</H3>A five digit number is the fourth power of an integer. The sum of the first, third, and fifth digits equals the sum of the second and fourth digits. What digit is in the thousands place ? <B>Solution</B> It is a nice problem to apply the divisibility rules. <pre> Since the sum of the first, third and fifth digits equals the sum of the second and fourth digits, it means that the alternate sum (or difference) of the digits is 0 (zero, ZERO). Then the rule of divisibility by 11 tells us that this 5-digit number is divisible by 11. Since 11 is a prime number, it implies that the 5-digit number is a multiple of 11^4, which is 14641. Among 5-digit numbers, there is ONLY ONE such number (4-th degree of a multiple of 11), since the next such number 22^4 = 234256 is just 6-digit number. So, only one number satisfies the problem's condition, and it is the number 14641. Its thousands digit is 4, which gives the answer to the problem's question. On the divisibility by 11 rule, see the lesson - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-11-rule.lesson>Divisibility by 11 rule</A> in this site. </pre> <H3>Problem 4</H3>Prove that for any given set of 17 integers, not all of which are odd, there exist nine of them whose sum is divisible by 2. <B>Solution</B> <pre> Let m be the number of odd integers in our sets of 17 integer numbers, and let n be the number of even integers in our sets of 17 integer numbers. We have m + n = 17. It implies that EITHER m >= 9 OR n >= 9. In other words, one of the two integer numbers, m or n, must be at least 9. Indeed, otherwise the sum m+n would not be more than 8 + 8 = 16; but m+n = 17. If n >= 9, then we can take 9 even integer numbers from our set. Their sum will be divisible by 2, so in this case these 9 integer numbers are the seeking set. If m >= 9, then there are two sub-cases: - (a) all integer numbers in our set are odd. It contradict to the imposed condition, so this case can not happen. - (b) there is at least one even number in our set of 17 integer numbers. In this case, we form the set of 9 numbers, taking 8 odd numbers and this even number. The sum of these 9 integer numbers is even number. So, we proved that in any case, it is possible to find a subset of 9 integers with even sum. </pre> The proof is complete and the problem is solved. My other lessons in this site on divisibility numbers are - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-2-rule.lesson>Divisibility by 2 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-3-rule.lesson>Divisibility by 3 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-4-rule.lesson>Divisibility by 4 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-5-rule.lesson>Divisibility by 5 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-6-rule.lesson>Divisibility by 6 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-9-rule.lesson>Divisibility by 9 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-10-rule.lesson>Divisibility by 10 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Divisibility-by-11-rule.lesson>Divisibility by 11 rule</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Restore-the-omitted-digit-in-a-number-in-a-way-the-number-is-divisible-by-9.lesson>Restore the omitted digit in a number in a way that the number is divisible by 9</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Restore-the-omitted-digit-in-a-number-in-a-way-the-number-is-divisible-by-11.lesson>Restore the omitted digit in a number in a way that the number is divisible by 11</A> - <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Can-there-be-a-perfect-square.lesson>Can there be a perfect square ?</A> - <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Math-circle-level-problem-on-restoring-digit-in-the-product-of-two-16-digit-numbers.lesson>Math circle level problem on restoring digit in the product of two 16-digit numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Math-circle-level-problem-on-finding-remainders.lesson>Math circle level problem on finding remainders</A> - <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/OVERVIEW-of-Divisibility-rules-by-2-3-4-5-6-9-10-11.lesson>OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11</A>
