TRANSNATIONAL LEARNING NETWORK COMENIUS PROGRAMME

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Lancaster's Maths Problems' SOLUTIONS November 2003

*********2003

Réponse de Cécile PERRIN , seconde 4. Lycée "en forêt", MONTARGIS
1. Pour le dernier chiffre :
Le dernier chiffre des puissances de 2 est soit 2, 4, 8, ou 6, dans cet ordre.
Donc tous les quatre nombres le même chiffre revient.
Or 2003 = 4 x 500 +3, donc la suite (2,4,8,6) est répétée 500 fois et recommence jusqu'au troisième chiffre de celle-ci soit 8
Conclusion : le dernier chiffre de 22003 est 8
2. Pour l'avant dernier chiffre :
Dans la suite des puissances de 2 lorsque le 8 est le dernier chiffre, l'avant dernier est tour à tour 0, 2, 4, 6 ou 8, donc tous les 5 nombres le même chiffre revient.
Or dans 22003 le 8 comme dernier chiffre revient 501 fois ( 500 fois plus une fois )
501 = 5 x 100 + 1 donc la suite des 5 chiffres ( 0,2,4,6,8) est répétée 100 fois et recommence au premier chiffre, soit 0
Conclusion : l'avant -dernier chiffre de 22003 est 0

Hi

The solution of your task is the following:
the last 2 digits of 2°2003 are 08, because the last digit repeats itself after 4 times (the order is 2,4,8,6) and the secondlast digit repeats itself after 22 times (here the order is 0,0,1,3,6,2,5,1,2,4,9,9,8,6,3,7,4,8,7,5).
if you now devide 2003 through 4, you get x.75 so the last digit is 8.
if you devide 2002 through 22 you get 91, so the secondlast digit has to be the second one from the order, so it has to be the 0.

bye bye from herderschool germany--

Maths Contest sub-project- Concours
Mathématiques- Mathematik- Wettbewerbe
November 2003 Maths Problem from Lancaster

2. Explain briefly how you found your answer.

22003=24x500+3=24x500x8=(24)500x8=16500x8
Now we have to find the last two digits of 16500x8
The last 2 digits of any number I abridge as u2d
u2d of (161)=16
u2d of (162)=56
u2d of (163)=96 we observe that the last 2 digits possible
u2d of(164)=36 for a power of 16 are 16, 56, 96, 36, 76
u2d of(165)=76
u2d of(166)=16
u2d of(167)=56
Now we divide 500(from16500) with the number of possible last 2 digits for a power of 16 500:5=100 We see that the rest of this division is 0(zero) the last 2 digits of 16500=76 because when we divide the power's exponent with the number of the possible last digit or last 2 digits or last n digits and the rest=1 then the last digit or last 2digits or last n digits is (are)the first possible last digit, 2 or n digits , when the rest=2 then the last digit or last 2 digits or the last n digits is(are) the second possible last digit, last 2 digits or last n digits and so on. When the rest=0 then the last digit, the last 2 digits or the last n digits is(are) the last possible last digit, last 2 digits or last n digits.("n" included in the multitude of natural numbers) u2d of (16500)=76
u2d of(16500x8)=u2d of(16500)x8=u2d of(76x8)=08
u2d of(16500x8)=08 u2d of(22003)=08 because 16500x8=22003 THE LAST TWO DIGITS OF 22003 ARE 08(the last digit is 8 and the penultimate digit is 0)

This exercise was solved and written by me, Filote Lucian from "Colegiul National Mihai Viteazul", Ploiesti, 6M classroom, Prahova, Romania

To solve this problem you will need to carry out an investigation.
1.What are the last two digits in the number whose value is :2 raised to the power 2003(Two multiplied by itself 2003 times)?
2²ºº³:2²=2²ºº¹Û2² | 22003(1)
A number is divisibil by fourÛits last two digits make a number divising by four.(2)
u(2²ºº³)=8(3)
u(2¹)=2(4)
u(2²)=4(5)
u(2³) =8(6)
u(24) =6(7)
u(25) =2(8)
From (1),(2),(3),(4),(5),(6),(7),(8)Þthe last two digits in 2²ºº³ can be: 0 and 8; 2 and 8;4 and 8;6 and 8;8 and 8.
The two last digits in 22004 are 1 and 6.(9)
22004 :2=2²ºº³(10)
24 :2=2³(11)
From (9),(10),(11)Þthe last two digits in 2²ºº³ are 0 and 8.

Nicolae Vlad-Mihai
CNMV Ploiesti
Sixth form

Romania

1. The two last digits in the number whose value is: 2 raised to the power 2003 are 08.

Les deux derniers chiffres dans le nombre 2^2003 sont 08.

2. We can notice a periodicity of 20 numbers starting with 2² :
04-08-16-32-64-28-56-12-24-48-96-92-84-68-36-72-44-88-76-52;
then, we do it again in the same order, the same numbers.
Because,
2^3 = 8
2^23 = 8388608
2^103 = 10141204801825835211973625643008
Therefore,
2^2003=....08.

We have also found the number 2^2003 on a TI89 so it is :

91850455621940361938626656094214558721785416167095616038211418946061300911389625108532758905320501593475

67717111899702187576951686904442651407450849333225483130399131624982400055102331859203589714314159205078

89958252923709171106911578536208323730831767239972465299190471557036844986493196289313775762661450753351

27187566804481903587846331247120747821200343202173365300664417606505789025033409014306653765279196317134

50161742388672323372788254660123271154712167353644396622687173763132806556577330446620430434046698081242

26790291355456436071322206271061001403484230406582742166543723630097478809192235008

On peut remarquer une suite de période 20 en commencant par 2² :
04-08-16-32-64-28-56-12-24-48-96-92-84-68-36-72-44-88-76-52;
puis, on recommence dans le même ordre, les mêmes nombres.
Car,
2^3 = 8
2^23 = 8388608
2^103 = 10141204801825835211973625643008
Donc,
2^2003=....08.

Nous avons aussi trouvé 2^2003 sur une TI89, ce qui nous donne donc:

9185045562194036193862665609421455872178541616709561603821141894606130091138962510853275890532050159347567

717111899702187576951686904442651407450849333225483130399131624982400055102331859203589714314159205078899582529237

0917110691157853620832373083176723997246529919047155703684498649319628931377576266145075335127187566804481903

58784633124712074782120034320217336530066441760650578902503340901430665376527919631713450161742388672323372788254

66012327115471216735364439662268717376313280655657733044662043043404669808124226790291355456436071322206271061001

403484230406582742166543723630097478809192235008

CLAVAL Elsa
SIMOES Garrick
LEROY Charlotte
DANNENMULLER Chloé
TermL1
Lycée en Forêt, Montargis, France