Some
multiplication [product] short cuts or tips:
1]
multiplication of the numbers with 5 in ones place;
1000's and 100's places 10's 1's places
A] Squares: eg: 1] 35X35=1225 [by
3x[3+1] 5x5]]
2] 75X75=5625 [by 7x[7+1] 5x5]]
*For
n5xn5=n[n+1]25 [n[n+1]is in 1000’s and 100’s place +25 is in 10’s and one’s
place]
B] the numbers with 5 in one’s place and differ by 10
Eg:1] 35X45=1575
2] 65x55=3575
*For [n-1]5xn5=[n2-1] 75
2]the numbers multiplied by 5
E.g1] 256x5=1280 [i.e [256/2]x10
E.g2] 445x5=2225 [i.e [445/2]x10+5]]
If an even number n multiplied by 5 the product is
[n/2]x10
If odd number n multiplied by 5 the product is [[n-1]/2]x10+5
3] square of the numbers near to 10,100,100…etc.
10,s one’s
E.g.1] 9x9=81 [9-1] 1
=81
100’s one’s
Eg. 2] 98x98= [[98-2] [100-98]x[100-98]=9604
i.e 98-[100-98] is
in 1000’s and 100’s place further [100-98][100-98] is in 10’s place and one’s
place.
Same way 96x94=[96-[100-94]
[100-94]x[100-96]]=9024.
3] product of the numbers near to 10,100,100…etc.
E.g1] [two places left to
10’s] two places left from start
1000’s, 100’s 10’s
1’s
Eg. 98x97=9506 [98-[100-97]] [100-98]x[100-97] as the
number of digits in every number is ‘’2’’
E.g2] Can you imagine the product of the two numbers
99994x99997
5places 5
places
Simple: 99994-[1000000-99997] [1000000-99994]x[100000-99997]=9999100018.
The number of digits in product=2xthe number of digits
in each number.
So we have
obtained the product by multiplying the each number differences to 100000 is 6x3=18 it should be in first 5 digits so that is 00018. Next, the
difference between the one of the numbers to Nearest Round number
100000-another number]
Product of any two numbers which are near to 50,
500, 5000 etc
Two places
Two places
E.g1:
47X48=[47-[50-48]]/2 [50-47]x[50-48]
here the left side number should be multiplied by 50/100 i.e ½
=[47-2]/2 3x2
=45/2 06
=22.5 06
=2256
The fraction 0.5
in 100,s place= direct number 5 in 10’s place [0.5x10=5]
Two places
Two places
Eg2] 44x46=
[44-[50-46]]/2 [50-44]x[50-46]
=[44-4]/2 6x4
= 40/2 24
=20 24
The product=2024
Three places three places
E.g3]
498x493=[498-[500-493]]/2 [500-498]x[500-493]
=[498-7]/2 2x7
= 491/2 014
=245.5 014
=245514
The fraction 0.5
in 1000’s place =direct integer in 100’s place.
Product of any two numbers which are near to 25, 250,
2500 etc
Two places
Two places
Eg1] 23x24=
[23-[25-24]]/4 [25-23]x[25-24] , the left side number should be multiplied by
50/100 i.e ¼
e.g2] = [23-1]/4 2x1
=22/4 02 as two places
=5.5 02
=552
The fraction 0.5
in 100’s place = direct integer 5 in 10’s place.
The all remaining product types are to be practiced as
the tips applied for some special cases to easy to solve.
The above particular cases only, the multiplication
tips applied.
For remaining, the normal method of practice is sufficient.
3] The 9 is a magic number can remember 9th
table easily.
A] 9th Table: Ten’s
place one’s place
9x1=9 0 [9-0]
9x2=18 1 [9-1]
9x3=27 2 [9-2]
9x4=36 3 [9-4]
9x5=45 4 [9-4]
9x6=54 5 [9-5]
9x7=63 6 [9-6]
9x8=72 7 [9-7]
9x9=81 8 [9-8]
9x10=90 9 [9-9]
10's place 1's place
9xn= [n-1] [10-n]
If we/ remind the 9 table up to 9x10=90 we
can solve the many multiplications by 9
The
remainder when a number divided by 9 with out division
1] 95÷9 quotient is 10 and remainder 5
2]
782÷9 quotient is 86 and remainder 8
In
each event we find out quotient for finding out remainder.
But
we can find out remainder individually just by adding the digits of the number
up to 1digit.
E.g.
78932185÷9 then remainder 7+8+9+3+2+1+8+5=43 then 4+3=7 so the remainder is 7.
By
eliminating the 9,s from sum obtained ≥ 9 while calculating sum, we can
get remainder easily
[7+8]-9+9-9+[3+2+1+8-9]+5=6+5+5
again-9=7
Add
upto 9 and eliminate
6+[1+8]+9+[3+2+[1+8]]+5
after eliminating the red colour sums we get
=6+5+5=[6+3]+2+5 again after eliminating the red colour
sums we get =7
2] By The same way if
a number divided by 3, we can find the remainder, without knowing
quotient.
E.g 57÷9,
the remainder is 5+7=12, further 1+2=3 but the 3 is divisible by 3 so remainder
is 0
E.g
43÷9 , the remainder is 4+3=7, the 7 is
not divisible by 3, so remainder is 7-3x2=1
All the best wishes for competitive examination and
aware of mathematics. [ a well known subject means ‘’can be expressed the
subject in mathematics’’]
[Next we approach to provide best understanding on divisibility and number
systems i.e types of numbers]
1]
multiplication of the numbers with 5 in ones place;
1000's and 100's places 10's 1's places
A] Squares: eg: 1] 35X35=1225 [by
3x[3+1] 5x5]]
2] 75X75=5625 [by 7x[7+1] 5x5]]
*For
n5xn5=n[n+1]25 [n[n+1]is in 1000’s and 100’s place +25 is in 10’s and one’s
place]
B] the numbers with 5 in one’s place and differ by 10
Eg:1] 35X45=1575
2] 65x55=3575
*For [n-1]5xn5=[n2-1] 75
2]the numbers multiplied by 5
E.g1] 256x5=1280 [i.e [256/2]x10
E.g2] 445x5=2225 [i.e [445/2]x10+5]]
If an even number n multiplied by 5 the product is
[n/2]x10
If odd number n multiplied by 5 the product is [[n-1]/2]x10+5
3] square of the numbers near to 10,100,100…etc.
10,s one’s
E.g.1] 9x9=81 [9-1] 1
=81
100’s one’s
Eg. 2] 98x98= [[98-2] [100-98]x[100-98]=9604
i.e 98-[100-98] is
in 1000’s and 100’s place further [100-98][100-98] is in 10’s place and one’s
place.
Same way 96x94=[96-[100-94]
[100-94]x[100-96]]=9024.
3] product of the numbers near to 10,100,100…etc.
E.g1] [two places left to
10’s] two places left from start
1000’s, 100’s 10’s
1’s
Eg. 98x97=9506 [98-[100-97]] [100-98]x[100-97] as the
number of digits in every number is ‘’2’’
E.g2] Can you imagine the product of the two numbers
99994x99997
5places 5
places
Simple: 99994-[1000000-99997] [1000000-99994]x[100000-99997]=9999100018.
The number of digits in product=2xthe number of digits
in each number.
So we have
obtained the product by multiplying the each number differences to 100000 is 6x3=18 it should be in first 5 digits so that is 00018. Next, the
difference between the one of the numbers to Nearest Round number
100000-another number]
Product of any two numbers which are near to 50,
500, 5000 etc
Two places
Two places
E.g1:
47X48=[47-[50-48]]/2 [50-47]x[50-48]
here the left side number should be multiplied by 50/100 i.e ½
=[47-2]/2 3x2
=45/2 06
=22.5 06
=2256
The fraction 0.5
in 100,s place= direct number 5 in 10’s place [0.5x10=5]
Two places
Two places
Eg2] 44x46=
[44-[50-46]]/2 [50-44]x[50-46]
=[44-4]/2 6x4
= 40/2 24
=20 24
The product=2024
Three places three places
E.g3]
498x493=[498-[500-493]]/2 [500-498]x[500-493]
=[498-7]/2 2x7
= 491/2 014
=245.5 014
=245514
The fraction 0.5
in 1000’s place =direct integer in 100’s place.
Product of any two numbers which are near to 25, 250,
2500 etc
Two places
Two places
Eg1] 23x24=
[23-[25-24]]/4 [25-23]x[25-24] , the left side number should be multiplied by
50/100 i.e ¼
e.g2] = [23-1]/4 2x1
=22/4 02 as two places
=5.5 02
=552
The fraction 0.5
in 100’s place = direct integer 5 in 10’s place.
The all remaining product types are to be practiced as
the tips applied for some special cases to easy to solve.
The above particular cases only, the multiplication
tips applied.
For remaining, the normal method of practice is sufficient.
3] The 9 is a magic number can remember 9th
table easily.
A] 9th Table: Ten’s
place one’s place
9x1=9 0 [9-0]
9x2=18 1 [9-1]
9x3=27 2 [9-2]
9x4=36 3 [9-4]
9x5=45 4 [9-4]
9x6=54 5 [9-5]
9x7=63 6 [9-6]
9x8=72 7 [9-7]
9x9=81 8 [9-8]
9x10=90 9 [9-9]
10's place 1's place
9xn= [n-1] [10-n]
If we/ remind the 9 table up to 9x10=90 we
can solve the many multiplications by 9
The
remainder when a number divided by 9 with out division
1] 95÷9 quotient is 10 and remainder 5
2]
782÷9 quotient is 86 and remainder 8
In
each event we find out quotient for finding out remainder.
But
we can find out remainder individually just by adding the digits of the number
up to 1digit.
E.g.
78932185÷9 then remainder 7+8+9+3+2+1+8+5=43 then 4+3=7 so the remainder is 7.
By
eliminating the 9,s from sum obtained ≥ 9 while calculating sum, we can
get remainder easily
[7+8]-9+9-9+[3+2+1+8-9]+5=6+5+5
again-9=7
Add
upto 9 and eliminate
6+[1+8]+9+[3+2+[1+8]]+5
after eliminating the red colour sums we get
=6+5+5=[6+3]+2+5 again after eliminating the red colour
sums we get =7
2] By The same way if
a number divided by 3, we can find the remainder, without knowing
quotient.
E.g 57÷9,
the remainder is 5+7=12, further 1+2=3 but the 3 is divisible by 3 so remainder
is 0
E.g
43÷9 , the remainder is 4+3=7, the 7 is
not divisible by 3, so remainder is 7-3x2=1
All the best wishes for competitive examination and
aware of mathematics. [ a well known subject means ‘’can be expressed the
subject in mathematics’’]
[Next we approach to provide best understanding on divisibility and number
systems i.e types of numbers]
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