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(1) {3,4=5]2 is the smallest Pythagoras
triple. What is the next higher Pythagoras triple having same ending digits ?
(3) 9 +9 = 18 ; 9 x9 = 81 Here the sum and the product of two numbers give results where the digits are revered. Can you find out more examples?
(4) x5 + x4 + x3 + x2
+ x + 1= 0. Find the value of x.
(7) Under what condition the relation (a+b)/(c+d) = (a2
– b2)/(c2 – d2) is true. Give numerical
example
(9) If x2
+ 1/x2 = 34 . Find the value of x5 + 1/x5
(1) [13,84=85]2
13 is prime;132
= (c-b)(c+b) ,if c-b=1 and c+b = 169 ,
c = 85 and b = 84
(2) (2n-1) + (2n+1) + (2n+3) = 6n +3 = 3 (2n+1). To give odd cube (2n+1) must be odd or (2n+1) = 9, 243 ------9 (2x-1)3 . It gives
7+9+11=27
= 3x3x3 = 33
241
+ 243 + 245 = 729 = 9x9x9 = 93
(3)
Let m and n be the two numbers satisfying the required condition. m+ n = 10 x +
y and mn = 10y + x where x, y are any single digit number.
m = 10x + y – n = (10y +x
)/n
which gives n2 = x(10n
-1)- y(10-n) To make n to be positive
x(10n-1) > y(10-n)
n=1; 1 = 9(x-y) , there is now
whole number solution n = 2; 4 = 19x – 8y, which gives x =4 and y =9
,or m = 47 2 + 47 = 49; 2 x 47 = 94
n=3 ; 9 = 29x –
7y, which provides x =2. Y =7 or m = 24 3 + 24 = 27 ; 3 x 24 = 72 For
higher values of n , n = 4,5,6… ., there is no such pair of numbers . If the sum and product are three digit numbers, then m+n = 100x + 10 y + z and mn = 100z + 10y +x m = (100z +10y +x)/n = 110x + 10y +z – n n2 = x(100n -1) +10y (n-1) – z(100-n) when n = 2 ; 4 = 199x + 10y – 98z when x = 4 , z= 9 we have 786 +
19y – 882 which gives y = 9 0r m –
497 2 +
497 = 499 ; 2 x 497 = 0=994 This is true for any number of 9 in between 4 and 7. 2 + 4997 = 4999 ; 2 x 4997 = 9994 2+49997 = 40000 ; 2 x 49997 = 99994
(4)
The given expression can be written as x (x4 + x3 + x2
+ x + 1) = -1. There are two possibilities.
x
= 1 and (x4 + x3 + x2 + x + 1) = -1 ; x = -1 and (x4 + x3 + x2 + x + 1) = 1
The
former solution is not valid as (x4 + x3 + x2 +
x + 1) ≠ -1 if x =1. But x = -1, satisfies both the conditions. The solution is
same to all problems where the exponents of x begin from 0 to (2n-1)
x3
+ x2 + x + 1 = 0
x7
+ x6 + x5 + x4 + x3 + x2
+ x +1 =0
(5) xyz + xyz +
xyz = zzz
3 (100 x + 10 y+
z) = 100 z + 10 z + z = 111 z
100x +10y + z = 37 z -à 100x +10y = 36
z à 50 x + 5 y = 18 z
The
digital roots must be same in both sides. Whatever may be the digital root of
z, because of its coefficient 36 , 36z will have a digital root 9. Hence the
digital root of x +y must be 9.
One possible
solution is z =5 , x= 1 and y =8)185, which gives 185 x 3 = 555
xyz + xyz + xyz =
yyy
3 ( 100 x + 10 y+
z) = 100 y + 10 y + y = 111 y
100 x + 10y + z = 37 y --à 100x + z = 27 y
The
digital roots must be same in both sides. Whatever may be the digital root of
y, because of its coefficient 27, 27 y
will have a digital root 9. Hence the digital root of x +z must be 9
One
possible solution is y =4 ,x=1 , z= 8
which gives 148 x 3 = 444
(
x≠w, as it gives forbidden solutions)
If the identical
digit (w) in the sum is different from the digits of the adder (x,y,z)
3(100x + 10y +z) =
100 w + 10w + w = 111 w
(100x + 10y +z) =
37 w
37 x 3n gives all
identical digits. Two solutions are available\
259 x 3 = 777 and
296 x 3 = 888
(6) We know a3 + b3 =
(a+b)(a2 – ab + b2 ) and a3 + c3 =
(a+c)(a2 – ac + c2 )
If
a2 – ab + b2 and a2 – ac + c2 are
equal (a+b)/ (a+c) =( a3 + b3
)/ (a3 + c3 )
The
condition demands a2 – ab + b2 = a2 – ac + c2
a(c-b) = c2 – b2 =
(c-b) (c+b) or a = c+b or c = a—bThe
general form of this relation is
(a
+b)/ [a +(a-b)] = (a3 + b3 )/ a3 + (a-b)3
. When a = 7 , b =5, we have (7+5)/ (7+ 2) = (73+53)/
(73+ 23)
(7)
(a+b)/(c+d) = (a2 – b2)/(c2 – d2) =
(a+b)(a-b) / (c+d)(c-d) .To make them to be equal a-b = c-d is the required
condition. For example, a=7,b=5 , a-b = 2, c = 3 , d = 1 and c-d =2, we have (7
+5)/(3+1) = 12/4 = (72 – 52)
/ (32 - 12 ) = 24/8
(8) Let us
suppose a + b = c+d and ab = cd.
Squaring the former relation a2 + b2 + 2ab = c2
+ d2 + 2cd and (a + b)3 = (c+d)3 -à a3 + b3 + 3ab ( a+b) = c3
+d3 + 3cd (c+d), Since ab = cd,
a2 + b2 = c2 + d2 and a3
+ b3 = c3 + d3
If a + b = c + d,
the difference between one pair (a-c) must be equal to the difference in the
other pair (d-b) That is if c = a +x ,
then d = b – x. Substitute these values in the relation ab = cd = (a+x)(b-x) =
ab + x(b-a) – x2 which gives
a condition x = b-a. Substituting this value in c = a + x = a + b – a = b and d
= b – x = b – b + a or d = b. The condition demands that two numbers must be
same in both sides of the relation. It proves that if sum of two numbers equals
to sum of two other numbers, then their product will not be equal, or if the
product of two numbers equals to product of two other numbers, then their sum
will not be equal.
(9) (x + 1/x)2
= x2 + 1/x2 + 2 = 34 + 2 = 36
Or x + 1/x = ± 6. With positive root, it gives a quadratic equation x2
– 6c + 1 = 0. Its solution is x = 3 ±2√ 2
(x
+ 1/x)3 = x3 + 1/x3 + 3(x + 1/x)
63 = 216 = x3 + 1/x3 + 18 or
x3 + 1/x3 =
216 – 18 = 198 (x2 + 1/x2)2 = (34)2 = 1156
= x4 + 1/x4 + 2 which gives x4 + 1/x4 = 1154 (x4 +
1/x4)(x + 1/x) = 1154 x 6 =6924 = (x5 + 1/x5)
+ (x3 + 1/x3)
which gives (x5 + 1/x5) = 6924 - (x3 + 1/x3) = 6924 –
198 = 6726
This can also be derived from (x + 1/x)5 (x + 1/x)5 =65
= 7776 = (x5 + 1/x5) + 5 (x3 + 1/x3)
+ 10 (x + 1/x)
(x5 + 1/x5) = 7776 – 5(198) – 10(6) =6726 Extending this idea (x6 + 1/x6) = (x3 + 1/x3)2 –
2 = 1982 -2 = 39202
(10) The general
form of the Pythagoras triple is (10n +8)2 + (10x +15)2 =
(10x + 17)2 which gives a condition 5n2 + 8n = 2x.For all
even n ,x becomes a positive number. When n = 2 , x = 18, which gives [28,195 =
197]2 and when n = 4 , x = 56 which gives [48,575=577]2
The general form of this set is (10n +8)2
+ (25n2 +40n +15)2 = (25n2 +40n + 17)2
