Continuing through the End of Chapter exercises, we encounter questions about multiples and divisibility.
Whenever you are asked to find numbers "divisible by $k$", you are instantly dealing with an Arithmetic Progression where the common difference $d$ is exactly $k$! The challenge lies in finding the exact starting number ($a$) and ending number ($t_n$) within the requested range before using your AP formulas.
Whenever you are asked to find numbers "divisible by $k$", you are instantly dealing with an Arithmetic Progression where the common difference $d$ is exactly $k$! The challenge lies in finding the exact starting number ($a$) and ending number ($t_n$) within the requested range before using your AP formulas.
1 Bounding Divisible Numbers
2 Summing Divisible Numbers
★ Complex Remainder APs
Select the correct option:
If you are creating a sequence of numbers that are "divisible by 8", what is the value of the common difference (\(d\))?
Concept:
1
The multiples of any integer $k$ form a perfect Arithmetic Progression where the gap between any two consecutive terms is exactly $k$.
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Hint:
1
Write out the first few multiples of 8: $8, 16, 24$.
2
Calculate $16 - 8$.
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Mistakes to Avoid:
!
Confusing multiples with powers: Numbers divisible by 8 grow by addition ($+8$), not by multiplication ($\times 8$).
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Solution:
For any sequence defined by divisibility by a factor, the common difference $d$ equals that factor. Hence $d = 8$.
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Select the correct option:
What is the smallest 3-digit number that is divisible by 6?
Concept:
1
To find a starting boundary ($a$), divide the minimum limit (100) by the divisor. Use the remainder to find the nearest valid multiple above the limit.
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Hint:
1
$100 \div 6 = 16$ with a remainder of 4.
2
This means $16 \times 6 = 96$. This is the largest 2-digit multiple.
3
Add 6 to 96 to enter the 3-digit range!
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Mistakes to Avoid:
!
Violating the digit constraint: Do not choose 96. Even though it is a multiple of 6 close to 100, the prompt explicitly requires a 3-digit number.
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Solution:
Next multiple after 96 ($16 \times 6$) is $17 \times 6 = 102$.
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Select the correct option:
How many multiples of 4 lie between 10 and 250?
Concept:
1
Identify the boundaries. Smallest multiple of 4 above 10 is 12 ($a=12$).
2
Largest multiple of 4 below 250 is 248 ($t_n=248$).
3
Solve $a + (n-1)d = t_n$ for $n$.
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Hint:
1
Equation: $12 + (n-1)4 = 248$.
2
$4(n-1) = 236$. Divide by 4.
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Mistakes to Avoid:
!
Assuming 250 is a multiple: 250 divided by 4 leaves a remainder of 2. You must subtract 2 to find the true upper boundary (248).
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Solution:
\(12 + (n-1)4 = 248\)
\((n-1)4 = 236\)
\(n-1 = 59 \implies n = 60\)
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Select the correct option:
What is the sum of the first 10 multiples of 5?
Concept:
1
The sequence is $5, 10, 15 \dots 50$. This is an AP where $a=5$, $l=50$, and $n=10$.
2
Use the finite AP sum formula: $S_n = \frac{n}{2}(a+l)$.
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Hint:
1
Calculate $\frac{10}{2} (5 + 50)$.
2
$5 \times 55 = ?$
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Mistakes to Avoid:
!
Just multiplying 10 by 5: 50 is only the *value* of the 10th term. The question asks for the SUM of all the terms leading up to it.
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Solution:
\(S_{10} = \frac{10}{2}(5 + 50)\)
\(S_{10} = 5 \times 55 = 275\)
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Select the correct option:
When solving for the number of terms ($n$) in divisibility problems, the final step usually looks like $(n-1) = X$. What is the critical final action?
Concept:
1
Because $t_n = a + (n-1)d$, solving algebraically isolates $(n-1)$. This $(n-1)$ physically represents the number of "jumps" taken from the start. To get the total count of positions, you must add the starting position back (+1).
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Hint:
1
If $(n-1) = 30$, then $n = 31$.
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Mistakes to Avoid:
!
The "Off-by-One" Error: The most common mistake in sequences is forgetting to add 1 at the end of the calculation, resulting in an answer that is exactly 1 unit too small.
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Solution:
To find $n$ from $n-1 = X$, you must definitively add $1$ to both sides, yielding $n = X + 1$.
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