Question

can you find the next no. having this property​

Answer 1

Here is your answer my friend—
The next number might be the double of 59 that is 118.
When 118 is divided by:
6, the remainder is 4.
5, the remainder is 3.
4, the remainder is 2.
3, the remainder is 1.



Hope it helps........
Please mark me as the brainliest!!!!!

Answer 2

So the number 59 has a property as mentioned in the question. These are due to the facts that,

• 59 - 1 = 58 is exactly divisible by 2.

• 59 - 2 = 57 is exactly divisible by 3.

• 59 - 3 = 56 is exactly divisible by 4.

• 59 - 4 = 55 is exactly divisible by 5.

• 59 - 5 = 54 is exactly divisible by 6.

Let us have a number 10a+b, where b is strictly a 1 - digit number and a is not strictly a 1 - digit number, can be of two or more, which can be written decimally as a/b, such that,

• a/(b-1) is exactly divisible by 2.

• a/(b-2) is exactly divisible by 3.

• a/(b-3) is exactly divisible by 4.

• a/(b-4) is exactly divisible by 5.

• a/(b-5) is exactly divisible by 6.

[Note:- Writing decimally means to write a number by splitting its digits by some symbols (here I used / which is not used for division.)]

Since a/(b-4) is exactly divisible by 5, the ones digit, i.e., b-4, must be either 0 or 5.

$\textit{``Multiples of 5 either ends in 0 or in 5 ."}$

Since the numbers a/(b-3) and a/(b-5) are exactly divisible by 4 and 6 respectively, they should be even, thereby implying that a/(b-4) should be odd.

Therefore, a/(b-4) ends in 5.

$\longrightarrow\sf{b-4=5}$

$\longrightarrow\sf{b=9}$

Hence our number is in the form a/9, such that,

• a/8 is exactly divisible by 2.      

• a/7 is exactly divisible by 3.

• a/6 is exactly divisible by 4.

• a/5 is exactly divisible by 5.

• a/4 is exactly divisible by 6.

Since a/6 is exactly divisible by 4, 'a' should be an odd number.

$\textit{``Tens digit of multiples of 4 ending in either 2 or 6 is always odd."}$

Tens digit of a/6 is the same as ones digit of 'a'. Since it is odd, so is 'a'. Let,

$\longrightarrow\sf{a=2x+1\quad\quad\dots(1)}$

for $\sf{x\in\mathbb{W}.}$

Since a/7 is exactly divisible by 3, sum of digits of 'a' added to 7 should be exactly divisible by 3.

Let sum of digits of 'a' be $\sf{S(a).}$ The sum of digits of a/7 will be $\sf{S(a)+7.}$

For a/7 being exactly divisible by 3,

$\longrightarrow\sf{S(a)+7\equiv0\pmod{3}}$

Since $\sf{7\equiv1\pmod3,}$

$\longrightarrow\sf{S(a)+1\equiv0\pmod{3}}$

$\longrightarrow\sf{S(a)\equiv-1\pmod{3}}$

Since $\sf{S(a)\equiv a\pmod{3},}$

$\longrightarrow\sf{a\equiv-1\pmod{3}}$

Then let,

$\longrightarrow\sf{a=3y-1\quad\quad\dots(2)}$

for $\sf{y\in\mathbb{N}.}$

From (1) and (2),

$\longrightarrow\sf{2x+1=3y-1}$

$\longrightarrow\sf{3y=2x+2}$

$\longrightarrow\sf{3y=2(x+1)}$

$\Longrightarrow\sf{2\,\mid\,3y}$

Since 2 is relatively prime to 3,

$\Longrightarrow\sf{2\,\mid\,y}$

Now we've to assume possible values for $\sf{n}$ to get possible numbers satisfying the property.

Let $\sf{y=2.}$ Then,

$\longrightarrow\sf{a=3\times2-1}$

$\longrightarrow\sf{a=5}$

Therefore,

$\longrightarrow\sf{a/9=59.}$

This is already described in the question.

Let $\sf{y=4.}$ Then,

$\longrightarrow\sf{a=3\times4-1}$

$\longrightarrow\sf{a=11}$

Therefore,

$\longrightarrow\sf{\underline{\underline{a/9=119}}}$

Thus 119 has the same property.

Let me find next possible number. Let $\sf{y=6.}$ Then,

$\longrightarrow\sf{a=3\times6-1}$

$\longrightarrow\sf{a=17}$

Therefore,

$\longrightarrow\sf{a/9=179}$

Thus 179 also has the same property.

Finally, let me find the general form of such numbers having the property.

For this we have to write a/9, written decimally, in general form as,

$\longrightarrow\sf{a/9=10a+9}$

From (2),

$\longrightarrow\sf{a/9=10(3y-1)+9}$

Since $\sf{2\,\mid\,y,}$ let $\sf{y=2n,\quad n\in\mathbb{N}.}$ Then,

$\longrightarrow\sf{a/9=10(3(2n)-1)+9}$

$\longrightarrow\sf{a/9=10(6n-1)+9}$

$\longrightarrow\sf{a/9=60n-10+9}$

$\longrightarrow\sf{\underline{\underline{a/9=60n-1}}}$

Therefore, numbers in the form $\sf{60n-1\quad\!\forall\,n\in\mathbb{N}}$ have the property mentioned in the question.