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$x$ is a positive integer less than 100 . When $x$ is divided by 5 , t
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Updated on: 08 Feb 2026, 08:39
Updated on: 08 Feb 2026, 08:39
Expert Reply
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Question Stats:
85% (02:23) correct
14% (02:28) wrong based on 7 sessions
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$x$ is a positive integer less than 100 . When $x$ is divided by 5 , the remainder is 4 , and when $x$ is divided by 23 , the remainder is 7 . What is the value of $x$ ?
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$x$ is a positive integer less than 100 . When $x$ is divided by 5 , t
[#permalink]
08 Feb 2026, 08:41
08 Feb 2026, 08:41
Expert Reply
To find the value of $x$, we can set up a system of linear congruences based on the information provided:
1. $\(x<100\)$
2. $\(x \equiv 4(\bmod 5)\)$
3. $\(x \equiv 7(\bmod 23)\)$
Step-by-Step Solution
1. Use the condition with the larger divisor first The second condition tells us that $x$ must be of the form:
$$
\(x=23 k+7\)
$$
where $k$ is an integer.
2. Test values of $k$ to satisfy $x<100$ We substitute different values for $k$ to find possible values for $x$, then check if they leave a remainder of 4 when divided by 5 :
- If $\(k=0: x=23(0)+7=7\)$.
- $\(7 \div 5=1\)$ with a remainder of 2 . (Incorrect)
- If $\(k=1: x=23(1)+7=30\)$.
- $\(30 \div 5=6\)$ with a remainder of 0 . (Incorrect)
- If $\(k=2: x=23(2)+7=53\)$.
- $\(53 \div 5=10\)$ with a remainder of 3 . (Incorrect)
- If $\(k=3: x=23(3)+7=76\)$.
- $\(76 \div 5=15\)$ with a remainder of 1 . (Incorrect)
- If $\(k=4: x=23(4)+7=99\)$.
- $\(99 \div 5=19\)$ with a remainder of 4 . (Correct!)
3. Verify the conditions
- Is 99 less than 100 ? Yes.
- Is $\(99 \div 5=19\)$ R 4 ? Yes.
- Is $\(99 \div 23=4 \mathrm{R} 7\)$ ? Yes $\((23 \times 4=92\)$, and $\(99-92=7)\)$.
Final Answer: The value of $x$ is 99 .
1. $\(x<100\)$
2. $\(x \equiv 4(\bmod 5)\)$
3. $\(x \equiv 7(\bmod 23)\)$
Step-by-Step Solution
1. Use the condition with the larger divisor first The second condition tells us that $x$ must be of the form:
$$
\(x=23 k+7\)
$$
where $k$ is an integer.
2. Test values of $k$ to satisfy $x<100$ We substitute different values for $k$ to find possible values for $x$, then check if they leave a remainder of 4 when divided by 5 :
- If $\(k=0: x=23(0)+7=7\)$.
- $\(7 \div 5=1\)$ with a remainder of 2 . (Incorrect)
- If $\(k=1: x=23(1)+7=30\)$.
- $\(30 \div 5=6\)$ with a remainder of 0 . (Incorrect)
- If $\(k=2: x=23(2)+7=53\)$.
- $\(53 \div 5=10\)$ with a remainder of 3 . (Incorrect)
- If $\(k=3: x=23(3)+7=76\)$.
- $\(76 \div 5=15\)$ with a remainder of 1 . (Incorrect)
- If $\(k=4: x=23(4)+7=99\)$.
- $\(99 \div 5=19\)$ with a remainder of 4 . (Correct!)
3. Verify the conditions
- Is 99 less than 100 ? Yes.
- Is $\(99 \div 5=19\)$ R 4 ? Yes.
- Is $\(99 \div 23=4 \mathrm{R} 7\)$ ? Yes $\((23 \times 4=92\)$, and $\(99-92=7)\)$.
Final Answer: The value of $x$ is 99 .
gmatclubot
$x$ is a positive integer less than 100 . When $x$ is divided by 5 , t [#permalink]
08 Feb 2026, 08:41
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