When a certain number is multiplied by 13, the product consists entirely of fives. The smallest such number is?

Let the number be x.
44² < x < 45²
? 1936 < x < 2025 ...(i)

From equation (i), the required number will be any number between 1936 and 2025. Since one part of the number is the square of 6 means one factor is 36.

? L, C, M of 36 and 5 = 180
? Number will be multiple of 180 i,e.,
180 x 11 = 1980 the only value which satisfies the equation (i).

Let first number = P
And second number = Q
According to the question
P - Q = 10 ....(i)
And 3P - 4Q = 18...(ii)

on multiply eq (i) by 3 and then subtract eq (ii) from it
3P - 3Q = 54
3P - 4Q = 18
Q = 36

On putting the value of Q in the eq (i)
P = 18 + Q = 18 +36
? P = 54
Required sum = P + Q = 54 +36 =90

Number of trees that can be planted on one side of road = (1760 / 20) + 1
= 88 + 1
=89

? Trees on the both side = 2 x 89=178

Let the third number = N
Then the first number = 3N
And second number = 3N/2
According to the question [N + 3N + 3N/2] / 3=154
? (2N +6N +3N) / 6 = 154
? N = (154 x 6) / 11 = 84
? Required difference = 3N - N = 2N
= 2 x 84 = 168

Let the number be P and Q
According to the question
(P + Q) = 2.5 (P - Q)
? P + Q = 2.5P - 25Q
? 3.5Q = 1.5P
? P / Q = 7/3 .....(i)
Now PQ = 84
? Q² = (84 x 3) / 7
? Q² = 36
? Q = 6
&ther4; P = (7/3) x 6 = 14

Sum of the number = P + Q = 14 + 6 = 20

let the two-digit number be 10p + q
Now according to the question
p + q = 10
and (p +10q) + 36 = (q + 10p)
? -9q + 9p = 36
? -q + p = 4 ....(ii)
on adding eq (i) and (ii) we get
2p = 14 ? p =7
? p = 7 and q = 4

? Required number is the 73
So, the number is a multiple of prime number.

The only one even prime number is 2 and this is not in the form of 3k + 1.
Thus any prime number of the form 3k + 1 is an odd number.
This means 3k must be even number which implies that k must be even number also.
Let k = 2m.
Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1.
Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

Let the two digit number be 10x + y.
Now, according to the question,
x + y = 10............(i)
if digits reversed , number get decreased by 36,
and ( x +10y) = ( 10x + y ) - 36
? - 9y + 9x = 36
? x - y = 4 ..................... (ii)
On adding Eqs. (i) and (ii), we get
2x = 14
? x = 7
? x = 7 and y = 3
? Required number is 73.
So,the number is a multiple of a prime number.

Let the fraction be x / y.
According to the question,
( x+2)/(y+1) = 1/2
? 2x + 4 = y+1
? 2x - y = -3 .........................(i)
Again according to the question,
(x+1)/(y-2) = 3/5
? 5x +5 = 3y - 6
? 5x - 3y = -11..........................(ii)
on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get,
multiplying Eq. (i) by 3
6x - 3y = -9
subtracting from Eq. (ii)
5x- 3y - ( 6x - 3y ) = -11 - (-9)
5x- 3y - 6x + 3y = -11 + 9
- x = - 2
x= 2
On putting the value of x in Eq. (i) we get
2x- y = -3
2 x 2 - y = -3
? 2 x 2 -y = -3
? -y = -3 - 4
? y = 7
? Original fraction = x/y =2/7

Method 1 to solve this question.
Let p be the first odd number.
Then next odd number is p + 2.
According to question,
Sum of five consecutive odd numbers 175
? p + p + 2 + p + 4 + p + 6 + p + 8 = 175
? 5p + 20 = 175
? p = (175 - 20)/5 = 31
According to question,
Sum of the second largest and square of smallest one = (31 + 6) + (31)²
Sum of the second largest and square of smallest one = 37 + 961 = 998

Method 2 to solve this question.
Let p be the even number.
Then next odd number is p + 1. so on p + 3 , p + 5................ P + 9
According to question,
Sum of five consecutive odd numbers 175
? p + 1 + p + 3 + p + 5 + p + 7 + p + 9 = 175
? 5p + 25 = 175
? p = (175 - 25)/5
? p = (150)/5
? p = 30
First odd number = p + 1 = 30 + 1 = 31
According to question,
Sum of the second largest and square of smallest one = (30 + 7) + (30 + 1)²
Sum of the second largest and square of smallest one = 37 + 961 = 998