@PURITAN said:remainder when 128^1000 is divided by 153 ??
52
@PURITAN said:remainder when 128^1000 is divided by 153 ??
153=17*9
E(17)=16
128^8 mod 17
9^8 mod 17
4^4 mod 17
52 mod 17
1
E(9)=6
128^4 mod 9
2^4 mod 9
7
17a+1=9b+7
17a=9b+6
a=3
Hence 52??
@techgeek2050 said:@Subhashdec2 y not (5,2,2) , (4,3,2) , (6,2,1)...?
432 will b considered but not 522 and 621 since they are not co prime to each other
so 6 more cases hence 15
@ravi.theja said:bhai it has to be 27 8c2-1 case...the case of (3,3,3) will make 39 as HCF..so its 27
i did not consider 333 case
@ravi.theja said:bhai it has to be 27 8c2-1 case...the case of (3,3,3) will make 39 as HCF..so its 27
it should b 15
711-3 cases
432-6 cases
531-6 cases
15 in total
@Subhashdec2 said:it should b 15 711-3 cases432-6 cases531-6 cases15 in total
2,2,5 -3 cases , 1,2,6 - 6 cases , 1,4,4 - 3 cases 😃 total 27
@PURITAN said:remainder when 128^1000 is divided by 153 ??
153 = 9*17
Now , use euler & chinese simultaneously to solve it.
Now , use euler & chinese simultaneously to solve it.
@naga25french said:Euler number of 17 is 16 .. so number formed by writing a unit digit 16 times consecutively will be divisible by 16 ..so 22222222 ( up to 32 times will be divisible by 17 and also 22222222 ( up to 16 times will be divisible by 171 - 32 --> digit divisible and 36 - 51 divisible by 17so we are left with 22x or x22 or 2x2Its now obvious .. Just check other posts for knowing what is euler number if you are aware of it .. hope this helps
so number formed by writing a unit digit 16 times consecutively will be divisible by 16 BUT HOW U DERIVED IT FOR 17??EULER IS APPLICABLE WHEN THE POWER IS GIVEN AS 16 OR MULTIPLE OF IT..PLS EXPLAIN..
@anuragu79 said:so number formed by writing a unit digit 16 times consecutively will be divisible by 16 BUT HOW U DERIVED IT FOR 17??EULER IS APPLICABLE WHEN THE POWER IS GIVEN AS 16 OR MULTIPLE OF IT..PLS EXPLAIN..
I already gave link to this theorem .. please check next page !
@ravi.theja said:The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
total 27 cases...
@ravi.theja said:The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
27
@ravi.theja said:The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
13x+13y+13z=117
x+y+z=9 =>11C2 =55 ordered solutions
What is the largest sum of rupees which can never be paid using infinite number of coins of denominations Rs. 5, Rs. 7 and Rs. 11?
x+y+z=9 =>11C2 =55 ordered solutions
What is the largest sum of rupees which can never be paid using infinite number of coins of denominations Rs. 5, Rs. 7 and Rs. 11?
@pirateiim478 said:13x+13y+13z=117x+y+z=9 =>11C2 =55 ordered solutionsWhat is the largest sum of rupees which can never be paid using infinite number of coins of denominations Rs. 5, Rs. 7 and Rs. 11?
13?
@pirateiim478 said:13x+13y+13z=117x+y+z=9 =>11C2 =55 ordered solutionsWhat is the largest sum of rupees which can never be paid using infinite number of coins of denominations Rs. 5, Rs. 7 and Rs. 11?
13?
@pirateiim478 said:13x+13y+13z=117x+y+z=9 =>11C2 =55 ordered solutionsWhat is the largest sum of rupees which can never be paid using infinite number of coins of denominations Rs. 5, Rs. 7 and Rs. 11?
13?? all the multiples of 5,7,11 can be paid and also...multiples of 12,18,16 can be paid..again...multiples of 17,19,31...can be paid........so it has to be 13??
@ravi.theja said:The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
13(a+b+c)=117
a+b+c=9
where a,b,c are coprime with each other
1,3,5
1,1,7
3 4 2
3 pairs
@ravi.theja said:The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
27??
13a+13b+13c=117
a+b+c=9
a,b,c>0
a+b+c=6
8c2
remove 1 case wen a=b=c=3
so
8c2-1
27??
13a+13b+13c=117
a+b+c=9
a,b,c>0
a+b+c=6
8c2
remove 1 case wen a=b=c=3
so
8c2-1
27??
@gnehagarg said:13(a+b+c)=117a+b+c=9where a,b,c are coprime with each other1,3,51,1,73 4 23 pairs
bhai its 27 luk at @chevy 's solution..xplained clearly