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Original question & working

Remove multiples and sum a bouncing path · source reference

[2014/NYJC/JC2/mid-year/P2/4] · Question pp. 4 · Working pp. 20, 21

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Printed question

(a) Find the sum of all integers in {1,2,3,…,2014}\{1,2,3,\ldots,2014\} that are not multiples of 3. [4]

(b) An elastic ball released from height hh falls vertically onto a horizontal floor and rebounds to height ahah, 0<a<10<a<1. Initially it is released from height HH.

(i) Write an expression in H,a,nH,a,n for its height after bounce nn. [1]

Given a=34a=\frac34,

(ii) find the number of bounces before the height is less than H/20H/20 for the first time [3];

(iii) calculate in terms of HH the total distance just before it comes to complete rest. [2]

Printed working

(a) Multiples of 3 are 3(1),3(2),…,3(n)3(1),3(2),\ldots,3(n) with 3n≤2014⇒n≤671133n\le2014\Rightarrow n\le671\frac13. There are 671 multiples.

Their sum =6712(3+2013)=676368=\frac{671}2(3+2013)=676368. Required sum =20142(1+2014)−676368=1352737=\frac{2014}2(1+2014)-676368=1352737.

(b)(i) Height after nnth bounce =anH=a^nH.

(ii) Given a=3/4a=3/4, consider

(3/4)nH<H/20⇒(3/4)n<1/20(H>0),(3/4)^nH<H/20\Rightarrow(3/4)^n<1/20\quad(H>0), nln⁡(3/4)<ln⁡(1/20)⇒n>ln⁡(1/20)ln⁡(3/4)>10.4.n\ln(3/4)<\ln(1/20)\Rightarrow n>\frac{\ln(1/20)}{\ln(3/4)}>10.4.

Least n=11n=11, so 11 bounces.

(iii) Total distance

H+2[(3/4)H+(3/4)2H+(3/4)3H+⋯ ]=H+2(3H/4)[1+3/4+(3/4)2+⋯ ]=H+32H11−3/4=7H.\begin{aligned}H+2[(3/4)H+(3/4)^2H+(3/4)^3H+\cdots]&=H+2(3H/4)[1+3/4+(3/4)^2+\cdots]\\&=H+\frac32H\frac1{1-3/4}=7H.\end{aligned}

Teacher notes

AP sum Sn=n2(u1+un)S_n=\frac n2(u_1+u_n) with n=671,u1=3,un=3(671)n=671,u_1=3,u_n=3(671).

Required integer sum = sum of all integers in {1,2,…,2014}\{1,2,\ldots,2014\} − sum of its multiples of 3.

Change inequality sign when dividing/multiplying by a negative value: ln⁡(3/4)<0\ln(3/4)<0.

Source diagrams redrawn

12-14

Printed answer key

(a) 1352737; (b)(i) anHa^nH; (ii) 11; (iii) 7H7H

Mathematical review and qualifications

(a) There are 671 multiples of 3. The required sum is 2014⋅2015/2−3⋅671⋅672/2=13527372014\cdot2015/2-3\cdot671\cdot672/2=1352737.

(b)(i) Rebound height is anHa^nH. (ii) (3/4)n<1/20(3/4)^n<1/20 gives n>ln⁡(1/20)/ln⁡(3/4)≈10.413n>\ln(1/20)/\ln(3/4)\approx10.413, so 11 bounces. (iii) Distance is H+2H∑n=1∞(3/4)n=H+2H(3)=7HH+2H\sum_{n=1}^{\infty}(3/4)^n=H+2H(3)=7H.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: ARITHMETIC & GEOMETRIC PROGRESSIONS. [2014/NYJC/JC2/mid-year/P2/4] Question PDF pages 4; solution PDF pages 20, 21. The original files remain private QA records outside this native content.