simple no. problem..
for what maximum value of n will the expression 10200!/504^n be an integer?
41^36) mod 7*11=?
(41^36) mod 7 = 1
(41^36) mod 11 = 3
so from here also remainder = 36
PLZ ANYONE EXPLAIN ME HW THIS 36 CAME...
7x + 1 = 11y + 3
least number satisfying this is 36
this is a methodical approach for these type of questions.
41^36 = 7k + 1 = 11m + 3
so 7k = 11m + 2
7k = 7m + (4m + 2)
k = m + (4m+2)/7
for integral k , 4m+ 2 = 7 p
so m = (7p - 2)/4 := p = 2
hence m = 3 , so 41^36 = 11m + 3 = 36
this is a methodical approach for these type of questions.
41^36 = 7k + 1 = 11m + 3
so 7k = 11m + 2
7k = 7m + (4m + 2)
k = m + (4m+2)/7
for integral k , 4m+ 2 = 7 p
so m = (7p - 2)/4 := p = 2
hence m = 3 , so 41^36 = 11m + 3 = 36
thanks for the answer, but i just wanted to know the logic behind this methodology...
simple no. problem..
for what maximum value of n will the expression 10200!/504^n be an integer?
As 504 is a composite number ( as it is not a prime number), we factor
it into its prime factors ,
504= 2^3 x 3^2 x 7
thus, we have to find the number of 7's that are contained in 10200!
(reason we look for number of 7's in this case is because number of 7's
here are less than the number of 2^3's and 3^2's , thus,as it is the
number of 7's that lays the restriction for the power of denominator,
we only have to find the number of 7's in 10200!)
the number of 7's in 10200! =
10200/7 + 10200/(7^2) + 10200/(7^3) +10200/(7^4)
= 1457 + 208 + 29 + 4 = 1698
How many natural number having at most five digits have the sum of their digits as at most five?
1) 216 2)428 3)286 4)251 5)512
Would really like to see the approach...
How many natural number having at most five digits have the sum of their digits as at most five?
1) 216 2)428 3)286 4)251 5)512
Would really like to see the approach...
9c4 + 8c4 + 7c4 + 6c4 + 5c4 = 251
@naga2french
cn u pl explain dat..
naga25french Says9c4 + 8c4 + 7c4 + 6c4 + 5c4 = 251
@naga2french
cn u pl explain dat..
a + b + c + d + e
number of sol = n + r - 1 C r - 1
so we have
a + b + c + d + e = 5------> 9C4
a + b + c + d + e = 4------> 8C4
a + b + c + d + e = 3------> 7C4
a + b + c + d + e = 2------> 6C4
a + b + c + d + e = 1------> 5C4
total ---------------------> 251
If logax1 > loga x2 then x1 > x2, if a > 1 and x1 x2, if a
How to prove this ?
Qualitative/Quantative / Examples ....whatever method ..does not matter .
Need a proof to believe the above fact.
Thanks for the time
How to prove this ?
Qualitative/Quantative / Examples ....whatever method ..does not matter .
Need a proof to believe the above fact.
Thanks for the time
If logax1 > loga x2 then x1 > x2, if a > 1 and x1 x2, if a
How to prove this ?
Qualitative/Quantative / Examples ....whatever method ..does not matter .
Need a proof to believe the above fact.
Thanks for the time
logax1 > loga x2 ===>
loga(x1/x2) > 0
log(x1/x2) / loga >0
if a log(x1/x2) must be -ve,==>
x1/x2 x1
if a > 1, loga > 0,==>log(x1/x2) must be +ve,==>
x1/x2 > 1 or x1 > x2
How many natural number having at most five digits have the sum of their digits as at most five?
1) 216 2)428 3)286 4)251 5)512
Would really like to see the approach...
a+b+c+d+e+f=5
Apply n+r-1Cr-1=10C5=252
Subtract one case when a,b,c,d,e each is equal to 0 and f=5
So total=251
i wud like 2 post a few probs which im nt able 2 solve...plzzzz help me out nd do post d reasonon or d soln plz dnt just post d answer....thnks n advance
1.find d remainder f 32^32^32/9
a.4 b.7 c.1 d.2
2. a similar one find remaindr f 50^51^52/11
a.6 b.4 c.7 d.3
3. remainder f 128^1000/153
a.103 b.145 c.118 d.52
4. remainde when 2^2 + 22^2 + 222^2 + 2222^2 +.......(2222...49twos)^2 divided by 9
a. 2 b. 5 c. 6 d. 7
5. Remainder when (1!)^3 + (2!)^3 + (3!)^3 + (4!)^3 +.....(1152!)^3 divided by 1152
a. 125 b. 225 c. 325 d. 205
plzzz help me out
1.find d remainder f 32^32^32/9
a.4 b.7 c.1 d.2
32^4k/9 --> 5^4k/9 --> 625/9 ------> 4
51^52 ends in 1
2. a similar one find remaindr f 50^51^52/11
a.6 b.4 c.7 d.3
E(11) = 10
so we have 50^10k+1 /11 --> 50/11 = 6
128^1000 = 2^7000
3. remainder f 128^1000/153
a.103 b.145 c.118 d.52
153 = 17 * 9
find 2^7000/ 17 & 2^7000/ 9
2^7000/ 17
E(17) = 16
2^16k+8/17 --> 256 /17 = 1
see the options
only 52 is of form 17k + 1 .. so 52
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2..
4. remainder when 2^2 + 22^2 + 222^2 + 2222^2 +......(2222...49twos)^2 divided by 9
a. 2 b. 5 c. 6 d. 7
find digit sum
N=2^2 + 4^2+ 6^2+..............98^2
=2^2(1^2+2^2+......49^2)
=4 * 49 * 50 * 99/6
=4*49*25*33
N mod 9
=4*4*7*6 mod 9
= -2*-3 mod 9
=6 mod 9
1152 = 2 * 24^2
5. Remainder when (1!)^3 + (2!)^3 + (3!)^3 + (4!)^3 +.....(1152!)^3 divided by 1152
a. 125 b. 225 c. 325 d. 205
4! = 24 .. so from 4!^3 the terms are perfectly divisble by 1152
so take only first three terms
(1!)^3 + (2!)^3 + (3!)^3 / 1152
(1 + 8 +216) /1152 = 225
how to find the number of non negative integral solution for x+2y+3z is there any formula/shortcut?
@naga25french
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2..
find digit sum
N=2^2 + 4^2+ 6^2+..............98^2
=2^2(1^2+2^2+......49^2)
=4 * 49 * 50 * 99/6
=4*49*25*33
N mod 9
=4*4*7*6 mod 9
= -2*-3 mod 9
=6 mod 9
hw did u gt d 2nd step where u write
N=2^2 + 4^2+ 6^2+..............98^2....i mean r u addin d no f 2's???hw s dat possible???its 2^2 + 22^2 + 222^2.......
i wud like 2 post a few probs which im nt able 2 solve...plzzzz help me out nd do post d reasonon or d soln plz dnt just post d answer....thnks n advance
1.find d remainder f 32^32^32/9
a.4 b.7 c.1 d.2
2. a similar one find remaindr f 50^51^52/11
a.6 b.4 c.7 d.3
3. remainder f 128^1000/153
a.103 b.145 c.118 d.52
4. remainde when 2^2 + 22^2 + 222^2 + 2222^2 +.......(2222...49twos)^2 divided by 9
a. 2 b. 5 c. 6 d. 7
5. Remainder when (1!)^3 + (2!)^3 + (3!)^3 + (4!)^3 +.....(1152!)^3 divided by 1152
a. 125 b. 225 c. 325 d. 205
plzzz help me out
1. (a) 4
2. (a) 6
3. (d) 52
4. remainde when 2^2 + 22^2 + 222^2 + 2222^2 +.......(2222...49twos)^2 divided by 9
= 2^2(1^2 + 11^2 + 111^2 + 1111^2 +............+ 1111.....49times^2) % 9
= 2^2(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 1^2 + 2^2 + 3^2...... 4times cyclic) % 9
= (4*4*(8*9*17)/6) % 9
= 16*204 % 9
= 6 (c)
5. (b) 225
explanations are clearly given by naga25french..
@naga25french
hw did u gt d 2nd step where u write
N=2^2 + 4^2+ 6^2+..............98^2....i mean r u addin d no f 2's???hw s dat possible???its 2^2 + 22^2 + 222^2.......
yea u can directly add if divisor is 9.. jreason is behind is divisibility rule of 9.. jus apply that ..
(22^33 +10^35)/45= find the rmndr
quite easy --but mah ans is not matching (courtesy nishit sinha )
(22^33 +10^35)/45= find the rmndr
quite easy --but mah ans is not matching (courtesy nishit sinha)
(22^33 +10^35)/45
45=5*9
22^33 = 2mod5
22^33 = 1mod9
5A+2=9B+1...37
10^35 = 0mod5
10^35 = 1mod9
5A+0=9B+1...10
(37+10)/45...2mod45
remandr=2
if 2 is remaindr givn in buk, thn u hav done sm mistake
if not thn buk(i m also following the same buk on sm1s recmndation,buk has many error....but dunno face ths problm)
if both are correct thn may be i m wrong.....thn correct me
