Weekly Challenge 6: AP train

7097 /www/nrich/html/content/id/7097/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/6426&part=">Try this next</a> </li> <li> <a href="http://nrich.maths.org/1333&part=">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/1418&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/7063&part=solution">Last week's solution</a> </li> </ul> <div> A particular list of $N$ consecutive integers starts with $1111$ as follows: $$1111, 1112, 1113, \dots, 1111 + N-1$$ The entire list is shifted $D$ places along the number line and the first number then excluded, leaving a list of $N-1$ larger consecutive numbers as follows: $$ 1112+D, 1113+D, \dots, 1111 + N -1+ D $$ In each list the sum of the integers is the same. What are the possibilities for $N$ and $D$? Extension: Maybe you wish to try to create a similar problem to this one? </div> <div class="framework"> Did you know ... ? Progressions of integers occur remarkably frequently in mathematics in applications from quantum mechanics to number theory and they have many beautiful properties. Even Carl Gauss, possibly the greatest mathematician of all time, fondly recalled his first encounter with sums of consecutive natural numbers, when he noticed that the sum of the first $100$ whole numbers  equalled $50$ lots of $101$.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This problem is based on the formula for the sum of the first $n$ natural numbers $$ 1+2+\dots + (n-1) + n =\frac{1}{2}n(n+1) $$ We have Its easier in a problem like this to introduce some notation. Let's write $S(n)$ to mean the sum of the first $n$ natural numbers. Since the two sums are equal we have $$ S(1111+N-1)-S(1110) = S(1111+N -1 +D) - S(1111 + D) $$ Let's put the formula into each of these and cancel each factor of a half. $$ (1110+N)(1111+N)-(1110)(1111)=(1110+N+D)(1111+N+D)-(1111+D)(1112+D) $$ Before leaping in we can see that many parts cancel, so we can put $$ 2221N + N^2 = 2221(N+D) + (N+D)^2 - 2223D - D^2- 2222 $$ Collecting things together gives $$ D(N-1) = 1111 $$ Now for a bit of number theory: $1111 = 11\times 101$. Thus, for solutions we require $$ D = 11, N=102\quad\mbox{ or } D=101, N=12 $$ You can see these two sums in action on this spreadsheet screenshot <mdo:image height="359" width="482" src="sums.png" alt=""></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This problem is based on the formula for the sum of the first $n$ natural numbers $$ 1+2+\dots + (n-1) + n =\frac{1}{2}n(n+1) $$ We have Its easier in a problem like this to introduce some notation. Let's write $S(n)$ to mean the sum of the first $n$ natural numbers. Since the two sums are equal we have $$ S(1111+N-1)-S(1110) = S(1111+N -1 +D) - S(1111 + D) $$ Let's put the formula into each of these and cancel each factor of a half. $$ (1110+N)(1111+N)-(1110)(1111)=(1110+N+D)(1111+N+D)-(1111+D)(1112+D) $$ Before leaping in we can see that many parts cancel, so we can put $$ 2221N + N^2 = 2221(N+D) + (N+D)^2 - 2223D - D^2- 2222 $$ Collecting things together gives $$ D(N-1) = 1111 $$ Now for a bit of number theory: $1111 = 11\times 101$. Thus, for solutions we require $$ D = 11, N=102\quad\mbox{ or } D=101, N=12 $$ You can see these two sums in action on this spreadsheet screenshot <mdo:image height="359" width="482" src="sums.png" alt=""></mdo:image> </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 6: AP train A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Sequences and Series AS Secondary Mapping Document DisplayCabinet