Number System Challenge – Win a FireUp T-Shirt
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FireUp brings to you a special challange, “Are You Smarter Than A FireUp Grad?”
Here are 6 special questions on Number system. Give the answers with proper solutions below to win a  Brand New Attractive FireUp T-Shirt!!!
1. Â Which is larger, 5^7 or 7^5?
2. Â Which is the largest three-digit number which when divided by 6 leaves the remainder 5, and when divided by 5 leaves the remainder 3?
3.  Prove that (244)^1500 – 1 is divisible by 1001. 1001 can be written in factorized form as 7 × 11 × 13.
4. Find the last digit of the number 
.
5. Â A garden planted saplings in such a way that every row had as many saplings as every column. If in all there were 729 trees, how many saplings were there in each row?
6. Â In the question no 5, if he decides to plant one new sapling between every two saplings, how many new saplings would have to plant?
Also mail your answers and contact details to info@fireup.co.in, so that we can send you the  prize.
The person with maximum correct answers with explanation will be given a special prize – A Brand New Attractive FireUp T-Shirt!!!
The answers and the winner(s) will be anounced on Monday, 31st August, 2009 at 1 PM.
August 26th, 2009 at 5:33 am
Answers to “Are You Smarter Than A FireUp Grad?†august 26 2009
1. 5^ 7 is greater
2. 983 is the greatest number.
4. 7
5. 27 samplings in each row.
6. totally 702 new samplings
August 26th, 2009 at 5:37 am
1. 5^7 > 7^5
2. 983
3. Can be verified by dividing 244^1500 individually by 7,11 and
13,each time we get a remainder 1,so 1-1 makes zero.
4. 7
5. 27
6. 53^2-27^2 = 2080
Forwarding the solutions to info@fireup.co.in
Regards,
Mayank Sharma
August 26th, 2009 at 6:57 am
hi ,
My answer
:
1)5^7
reason: 5^7 & 7^5=>5^7 & 7^(7-2)=>5^7 & 7^7/7^2=> 7^2 & (7/5)^7=> 49 & (1.4)^7=> 49>1.4^7=> 5^7>7^5
2)983:
let the smallest number satisfying given condition be N,
now we know n gives remainder 5 with 6 so this N can be represented as =>6*p + 5
similarly N when divided by 5 gives remainder 3
so N= 5q+3
solving we get the smallest number will be 23
now next number satisfying this condition would be 23 +(l.c.m 6,5) i.e 23+30==53
so largest three digit number would be 23 + 30*32=983
3)(244^1500-1)/7=>(245-1)^1500/7-1/7=> remainder is (-1)^1500/7-1/7=0
(244^1500-1)/11=>2^1500/11-1/11. Now according to eulers thm, 2^1500%11=1.hence, 2^1500-1 is divisible by 11
|||y we can prove tht (244^1500-1)/13 gives remainder 0.
hence, proved.
4)7
7^11^22^33=>7^(4m+1)(since, cyclicity of 7 is 4)
hence, unit digit will be 7.
5)27
x rows n x columns.
now x.x=729=> x=27
6)1404
for 3×3
0×0x0
x x x
0×0x0
x x x
0×0x0
12 new sapplings r required. generalising, we arrive at the following formula: 2n(n-1) new sapplings
so for n=27 ; 2*27*26=1404
August 26th, 2009 at 7:00 am
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August 26th, 2009 at 7:08 am
1. 5^ 7
2. 983
4. 7
5. 27
6.2080
I cannot give a better explaination than sanaksy. However, for last question he overlooked the fact tht even diagonally there will be new sapplings pasted.
so the formula will be 2*n*(n-1)+(n-1)^2
0×0x0
xxxxx
0×0x0
xxxxx
0×0x0
so ans is 2*26*27+26^2=>1404+676=2080
August 26th, 2009 at 2:05 pm
1.5^7=78125>7^5=16807
2.983
3.
4.7
5.27
6.729+(26*27)=1431
August 26th, 2009 at 2:23 pm
sent all the solutions
August 26th, 2009 at 5:29 pm
1)5^7 is larger
reason–since log is an increasing function…hence: 7log(5)>5log(7).
2)999/6=166+3;
therefore, 995-(6)k will be leaving remainder 5.
995/5=199;
therefore, 993-(5)k will be leaving remainder3.
common of the two will be my answer.
3)
4)n=7^(11*22*33)
=7^(7986)
therefore,the unit’s place will have:9
5)n=sqrt(729)=27 saplings.
6)n=26*27*2*2=2808.
August 27th, 2009 at 12:32 am
1. 5^7
2. 983
3. Can be verified by dividing 244^1500 individually by 7,11 and
13,each time we get a remainder 1,so 1-1 makes zero.
4. 7
5. 27
6. 53^2-27^2 = 2080
August 27th, 2009 at 12:46 am
1. 5^7>7^5
2. 983
3. 3)(244^1500-1)/7=>(245-1)^1500/7-1/7=> remainder is (-1)^1500/7-1/7=0
(244^1500-1)/11=>2^1500/11-1/11. Now according to eulers thm, 2^1500%11=1.hence, 2^1500-1 is divisible by 11
|||y we can prove tht (244^1500-1)/13 gives remainder 0.
hence, proved.
4.7
5.5)27
x rows n x columns.
now x.x=729=> x=27
6)1404
for 3×3
0×0×0
x x x
0×0×0
x x x
0×0×0
12 new sapplings r required. generalising, we arrive at the following formula: 2n(n-1) new sapplings
so for n=27 ; 2*27*26=1404
August 27th, 2009 at 2:15 am
1)5^7
reason: 5^7 & 7^5=>5^7 & 7^(7-2)=>5^7 & 7^7/7^2=> 7^2 & (7/5)^7=> 49 & (1.4)^7=> 49>1.4^7=> 5^7>7^5
2.983
It is nothing but 2 AP series with one first term is 5 and c.d of 6 and other first term is 3 and c.d is 5 we have to find largest common term less than 1000.
First common term is 23 and the next term is 56(23+lcm of 6&5) ans so on……
we get a new series of first term is 23 and c.d 30.
so,weget the ans 983
3. )(244^1500-1)/7=>(245-1)^1500/7-1/7=> remainder is (-1)^1500/7-1/7=0
(244^1500-1)/11=>2^1500/11-1/11. Now according to eulers thm, 2^1500%11=1.hence, 2^1500-1 is divisible by 11
|||y we can prove tht (244^1500-1)/13 gives remainder 0.
hence, proved.
4.7^11^22^33
we divide 11^22^33 by 4
and got the unit digit 1.
so unit digit of 7^11^22^33 = 7^(4k+1) = 7^1 =7.
5.N*N = 729
so N=27
6. 702.
since i check it by putting the value of N=5
we have to plant 20 new sapling.
means for N*N we need N*(N-1).
so we plant 27*26=702.
August 27th, 2009 at 2:47 am
1)it is 5^7
power of 5 is greater
2)as the last three digit no.is 999 we have to go through it
999/6 =3 (n.a)
995/6 =5 but divisible by 5 (n.a)
989/6 =5 but remainder 4 when divide by 5 (n.a)
983/6 =5 and when divide by 5 remainder 3 (applicable)
4)as 7 has 11 as its power always the last digit 7
5)x rows n x columns.
now x.x=729=> x=27
6)
for 3×3
0×0×0
x x x
0×0×0
x x x
0×0×0
12 new sapplings r required. generalising, we arrive at the following formula: 2n(n-1) new sapplings
so for n=27 ; 2*27*26=1404
3)3)(244^1500-1)/7=>(245-1)^1500/7-1/7=> remainder is (-1)^1500/7-1/7=0
(244^1500-1)/11=>2^1500/11-1/11. Now according to eulers thm, 2^1500%11=1.hence, 2^1500-1 is divisible by 11
|||y we can prove tht (244^1500-1)/13 gives remainder 0.
hence, proved.
August 27th, 2009 at 2:48 am
1)it is 5^7
power of 5 is greater
2)as the last three digit no.is 999 we have to go through it
999/6 =3 (n.a)
995/6 =5 but divisible by 5 (n.a)
989/6 =5 but remainder 4 when divide by 5 (n.a)
983/6 =5 and when divide by 5 remainder 3 (applicable)
4)as 7 has 11 as its power always the last digit 7
5)x rows n x columns.
now x.x=729=> x=27
6)
for 3×3
0×0×0
x x x
0×0×0
x x x
0×0×0
12 new sapplings r required. generalising, we arrive at the following formula: 2n(n-1) new sapplings
so for n=27 ; 2*27*26=1404
3)(244^1500-1)/7=>(245-1)^1500/7-1/7=> remainder is (-1)^1500/7-1/7=0
(244^1500-1)/11=>2^1500/11-1/11. Now according to eulers thm, 2^1500%11=1.hence, 2^1500-1 is divisible by 11
|||y we can prove tht (244^1500-1)/13 gives remainder 0.
hence, proved.
August 28th, 2009 at 5:39 am
hi,
sol (1) : 5^7 is greater.
explanation: we know that log(x) is an increasing function for x >0.so taking log both sides on LHS: 7 log5 = 7*0.69 (it could
easily seen the value would be ~ 7*.7 =4.9
now in RHS: 5 log7 = 5*log7 it is clear it would be less than 5log8 = 5*3 * log2 = 15* .303 = 4.5.
so easily ..it can be seen 5^7 is greater.
sol(2) :the required no is 983
explanation: we know that the largest 3 digit no.is 999.
now..the first no.is of form 6n+5 .it can easily seen it’s largest value would be 995.so we would have a decreasing AP of common
diff=6.so the different no’s are 995 986 983 977 ….
now the second no is of the form 5n+3.It’s largest value would be 998.So again it is a decreasing AP of C.D =5 so the diff. no’s
are 998 993 988 983 978…
so clearly the largest required no is 983.
sol(3):applying remainder theory:
we just need to prove that (244)^1500/1001 gives remainder 1.
now given 1001 = 7*11*13
so dividing 244^1500 by each no. as
244^1500 /7 = (245-1)^1500/7=(-1)^1500/7=remainder : 1.
NEXT: 244^1500 /11 = 2^1500/11 = (2^5)^300/11 = (33-1)^300/11 = (-1)^1500/7 =remainder : 1.
next:244^1500 /13 = 10^1500/13 = 9^750/13 = 81^375/13 = 3^375/13 = 27^125/13 = (26+1)^125/13= remainder : 1.
so clearly (244)^1500/1001 gives remainder 1 and hence proved [{(244)^1500}-1] is divisible by 1001.
sol(4)the unit digit is 7.
explanation: we have the exponents of the form 11^22^33 ..now taking 11^22^33 ..it wiil be of 4m+1 form clearly.
so we get 7^(4m+1) and it will give the unit digit 7.(the cyclic order(4) for 7 :7,9,3,1)
sol(5) there were 27 saplings in each row.
explanation : there is the given matrix n*n form (row*coloumn)..since no.of row = no.of coloumn.
total sapling would be n^2, which is given 729.
so clearly n = 27.
sol(6) there would be 1404 new saplings.
explanation : in first row for n sapling ..new sapling would be (n-1) for n even or odd,as per given condition.
in total n rows the no. of new saplings is n*(n-1).
now between every two adjacent rows no. of new saplings is n.In total = n*(n-1) since there is n-1 adjacent rows.
over all no. of new saplings is =2n*(n-1).
here n = 27..hence no. of new saplings is = 2*27*26 =1404
August 29th, 2009 at 4:36 am
1. Which is larger, 5^7 or 7^5?
A In this 5^7 is larger than 7^5 because 5^7=78125 and 7^5=16807.
2. Which is the largest three-digit number which when divided by 6 leaves the remainder 5, and when divided by 5 leaves the remainder 3?
A The largest three digit number is 983 which when divided by 6 leaves the remainder 5, and when divided by 5 leaves the remainder 3.
Method:- 6(5x+3)5 = 30x+23
In three digit number largest value of x =32
So, (30*32) + 23 =983.
this (244)^1500 – 1 is not divisible by 1001.
Because is divisible by 7 but not by 11 and 13.
4 Find the last digit of the number .
A The last digit is 7.
Method:- 22^33 last digit is 2 ,so the last two digit in 11^22^33 = 21
When 21 is divided by 4, the remainder is 1.
So the last digit is 7.
5. A garden planted saplings in such a way that every row had as many saplings as every column. If in all there were729 trees, how many saplings were there in each row?
A 729 is the square root of 27.
So saplings in each row is 27 and in every row and in every column is 1.
6. In the question no 5, if he decides to plant one new sapling between every two saplings, how many new saplings would have to plant?
A Total 1346 saplings would have to plant between every two saplings.In row between every two saplings we have to plant 27*26 saplings and In column between every two saplings we have to plant 27*26 saplings.
So (27*26) + (27*26) = 1346
August 29th, 2009 at 4:42 am
1. Which is larger, 5^7 or 7^5?
A In this 5^7 is larger than 7^5 because 5^7=78125 and 7^5=16807.
2. Which is the largest three-digit number which when divided by 6 leaves the remainder 5, and when divided by 5 leaves the remainder 3?
A The largest three digit number is 983 which when divided by 6 leaves the remainder 5, and when divided by 5 leaves the remainder 3.
Method:- 6(5x+3)5 = 30x+23
In three digit number largest value of x =32
So, (30*32) + 23 =983.
this (244)^1500 – 1 is not divisible by 1001.
Because is divisible by 7 but not by 11 and 13.
4 Find the last digit of the number .
A The last digit is 7.
Method:- 22^33 last digit is 2 ,so the last two digit in 11^22^33 = 21
When 21 is divided by 4, the remainder is 1.
So the last digit is 7.
5. A garden planted saplings in such a way that every row had as many saplings as every column. If in all there were729 trees, how many saplings were there in each row?
A 729 is the square root of 27.
So saplings in each row is 27 and in every row and in every column is 1.
6. In the question no 5, if he decides to plant one new sapling between every two saplings, how many new saplings would have to plant?
A Total 1404 saplings would have to plant between every two saplings.In row between every two saplings we have to plant 27*26 saplings and In column between every two saplings we have to plant 27*26 saplings.
So (27*26) + (27*26) = 1404
August 29th, 2009 at 5:49 am
1.5^7 is greater
2.983
the max 3 digit no. i.e divisible by 6 is 996 so 995 gives
5 as remainder but by dividing with 5 rem is 0 so (995-6)
aiso doesn`t satisfies 995-12=983 satisfies the condition
4.7
power of 11 has always 1 as last digit.so an odd number
power of 7 an odd no of times gives the last no`s as 7&3
it gives 7 as last digit
5.27
It is like a square so,
let `x` be the no of trees in a row,then x*x=729
x=27
6.1404
Between 27 trees we can place 26 new trees in a row
so (26*27) similarly, along columnwise (26*27)
totally 2*(26*27)=1404
September 1st, 2009 at 3:47 am
winner name nnouced or not
September 1st, 2009 at 4:07 am
WINNERS:
1. Mayank Sharma
2. Shailendra Tyagi
Thanks friends for participating in this challenge.We have got a lot of mails with correct and incorrect answers. We have shortlisted these two above winners.So congratulations to them and best of luck for the rest.We will contact the winners very soon.