a 6 digit largest number which is divisible by 4 but not divisible by 3 then find the sum of the digits of the number?
Answers
Answer:
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Answer:
1233
Step-by-step explanation:
First let's find the sum of all 2 digit numbers which are divisible by 3.
The least 2 digit multiple of 3 is 12 and the greatest is 99.
Let the multiples are in an AP.
Then the AP will be 12, 15, 18,..., 96, 99.
First term = T_1T
1
= 12
n^{th}n
th
term = Last term = T_nT
n
= 99
Common difference d = 3
No. of terms = n =
99−12
+1
3
87
=1
29+1
30
n = 30
Sum of terms = S_nS
n
=
\begin{gathered}\frac{30}{2}[12 + 99] \\ \\ 15 \times 111 \\ \\ 1665\end{gathered}
2
30
[12+99]
15×111
1665
Now let's find the sum of all 2 digit numbers which are divisible by 12. Because these are the 2 digit numbers which are divisible by not only 3 but also 4. We're going to deduct this sum from 1665 to get the answer. So this question is very simple!!!
The least 2 digit multiple of 12 is 12 and the greatest is 96.
Let these are in an AP.
Then it will be 12, 24, 36,..., 84, 96.
d = 12
T_1T
1
= 12
T_nT
n
= 96
n =
\begin{gathered}\frac{96 - 12}{12} + 1 \\\\\frac{84}{12} + 1 \\\\7 + 1 \\\\8\end{gathered}
12
96−12
+1
12
84
+1
7+1
8
n = 8
S_nS
n
=
\begin{gathered}\frac{8}{2}[12 + 96] \\ \\ 4 \times 108 \\ \\ 432\end{gathered}
2
8
[12+96]
4×108
432
S 432
Now subtract this from 1665 to get the answer.
1665 - 432 = 1233
∴ 1233 is the answer.
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